{"id":1029,"date":"2023-04-30T10:00:07","date_gmt":"2023-04-30T04:30:07","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=1029"},"modified":"2023-05-01T09:38:50","modified_gmt":"2023-05-01T04:08:50","slug":"trigonometric-identities","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/trigonometric-identities\/","title":{"rendered":"Trigonometric Identities"},"content":{"rendered":"<h2><strong>Trigonometric Identities<\/strong><\/h2>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/KlM76clMlD4?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>(1) \\(\\tan \\theta =\\frac{\\sin \\theta }{\\cos \\theta }\\text{ (linear)}\\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-128091\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-1.jpg\" alt=\"Trigonometric Identities 1\" width=\"727\" height=\"479\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-1.jpg 727w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-1-300x198.jpg 300w\" sizes=\"(max-width: 727px) 100vw, 727px\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-128092\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-2.jpg\" alt=\"Trigonometric Identities 2\" width=\"557\" height=\"494\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-2.jpg 557w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-2-300x266.jpg 300w\" sizes=\"(max-width: 557px) 100vw, 557px\" \/><\/p>\n<h3>Conditional trigonometrical identities<\/h3>\n<p>We have certain trigonometric identities.<br \/>\nLike sin<sup>2<\/sup> \u03b8 + cos<sup>2<\/sup> \u03b8 = 1 and 1 + tan<sup>2<\/sup> \u03b8 = sec<sup>2<\/sup> \u03b8 etc.<br \/>\nSuch identities are identities in the sense that they hold for all value of the angles which satisfy the given condition among them and they are called conditional identities.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-128094\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-3.jpg\" alt=\"Trigonometric Identities 3\" width=\"492\" height=\"623\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-3.jpg 492w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-3-237x300.jpg 237w\" sizes=\"(max-width: 492px) 100vw, 492px\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-128096\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-4.jpg\" alt=\"Trigonometric Identities 4\" width=\"488\" height=\"438\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-4.jpg 488w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-4-300x269.jpg 300w\" sizes=\"(max-width: 488px) 100vw, 488px\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-128102\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-5.jpg\" alt=\"Trigonometric Identities 5\" width=\"453\" height=\"414\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-5.jpg 453w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2017\/04\/Trigonometric-Identities-5-300x274.jpg 300w\" sizes=\"(max-width: 453px) 100vw, 453px\" \/><\/p>\n<h2><strong>Trigonometric Identities With Examples<\/strong><\/h2>\n<p><strong>Example 1:<\/strong> \u00a0 \u00a0Prove the following trigonometric identities:<br \/>\n(i) (1 \u2013 sin<sup>2<\/sup>\u03b8) sec<sup>2<\/sup>\u03b8 = 1<br \/>\n(ii) cos<sup>2<\/sup>\u03b8 (1 + tan<sup>2<\/sup>\u03b8) = 1<br \/>\n<strong>Sol.\u00a0 \u00a0 (i)<\/strong>\u00a0 We have,<br \/>\nLHS = (1 \u2013 sin<sup>2<\/sup>\u03b8) sec<sup>2<\/sup>\u03b8<br \/>\n= cos<sup>2<\/sup>\u03b8 sec<sup>2<\/sup>\u03b8 \u00a0 \u00a0 \u00a0 \u00a0\u00a0[\u2235\u00a01 \u2013 sin<sup>2<\/sup>\u03b8 = cos<sup>2<\/sup>\u03b8]<br \/>\n\\( ={{\\cos }^{2}}\\theta \\left( \\frac{1}{{{\\cos }^{2}}\\theta } \\right)\\left[ \\because \\ \\ \\sec \\theta =\\frac{1}{\\cos \\theta } \\right]\\)<br \/>\n= 1 = RHS<br \/>\n<strong>(ii) \u00a0<\/strong>We have,<br \/>\nLHS = cos<sup>2<\/sup>\u03b8 (1 + tan<sup>2<\/sup>\u03b8)<br \/>\n= cos<sup>2<\/sup>\u03b8 sec<sup>2<\/sup>\u03b8 \u00a0 \u00a0 \u00a0 \u00a0\u00a0[\u2235\u00a01 + tan<sup>2<\/sup>\u03b8 = sec<sup>2<\/sup>\u03b8]<br \/>\n\\( ={{\\cos }^{2}}\\theta \\left( \\frac{1}{{{\\cos }^{2}}\\theta } \\right)\\left[ \\because \\ \\ \\sec \\theta =\\frac{1}{\\cos \\theta } \\right]\\)<br \/>\n= 1 = RHS<\/p>\n<p><strong>Example 2:<\/strong> \u00a0 \u00a0Prove the following trigonometric identities:<br \/>\n\\( (\\text{i})\\text{ }\\frac{\\sin \\theta }{1-\\cos \\theta }=\\text{cosec}\\theta +\\cot \\theta \\)<br \/>\n\\( (\\text{ii})\\text{ }\\frac{\\tan \\theta +\\sin \\theta }{\\tan \\theta -\\sin \\theta }=\\frac{\\sec \\theta +1}{\\sec \\theta -1} \\)<br \/>\n<strong>Sol.\u00a0 \u00a0 (i)<\/strong>\u00a0 We have,<br \/>\n\\( LHS=\\frac{\\sin \\theta }{(1-\\cos \\theta )}\\times \\frac{(1+\\cos \\theta )}{(1+\\cos \\theta )} \\)<br \/>\n[Multiplying numerator and denominator by (1 + cos\u03b8)]<br \/>\n\\( =\\frac{sin\\theta (1+cos\\theta )}{1co{{s}^{2}}\\theta }=\\frac{\\sin \\theta (1+\\cos \\theta )}{{{\\sin }^{2}}\\theta } \\)<br \/>\n[\u2235 1 \u2013 cos<sup>2<\/sup>\u03b8 = sin<sup>2<\/sup>\u03b8]<br \/>\n\\( =\\frac{1+\\cos \\theta }{\\sin \\theta }=\\frac{1}{\\sin \\theta }+\\frac{\\cos \\theta }{\\sin \\theta } \\)<br \/>\n= cosec\u03b8 + cot\u03b8 = RHS<br \/>\n\\( \\left[ \\because \\ \\ \\frac{1}{\\sin \\theta }=\\cos ec\\theta \\,\\,and\\,\\frac{\\cos \\theta }{\\sin \\theta }=\\cot \\theta \\right] \\)<br \/>\n<strong>(ii)<\/strong>\u00a0\u00a0We have,<br \/>\n\\( LHS=\\frac{\\tan \\theta +\\sin \\theta }{\\tan \\theta -\\sin \\theta } \\)<br \/>\n\\( \\frac{\\frac{\\sin \\theta }{\\cos \\theta }+\\sin \\theta }{\\frac{\\sin \\theta }{\\cos \\theta }-\\sin \\theta }=\\frac{\\sin \\theta \\left( \\frac{1}{\\cos \\theta }+1 \\right)}{\\sin \\theta \\left( \\frac{1}{\\cos \\theta }-1 \\right)} \u00a0\\)<br \/>\n\\( \\frac{\\frac{1}{\\cos \\theta }+1}{\\frac{1}{\\cos \\theta }-1}=\\frac{\\sec \\theta +1}{\\sec \\theta -1}=RHS \\)<\/p>\n<p><strong>Example 3:<\/strong> \u00a0 \u00a0Prove the following identities:<br \/>\n(i) (sin\u03b8 + cosec\u03b8)<sup>2<\/sup> + (cos\u03b8 + sec\u03b8)<sup>2<\/sup> = 7 + tan<sup>2<\/sup>\u03b8 + cot<sup>2<\/sup>\u03b8<br \/>\n(ii) (sin\u03b8 + sec\u03b8)<sup>2<\/sup> + (cos\u03b8 + cosec\u03b8)<sup>2<\/sup> = (1 + sec\u03b8 cosec\u03b8)<sup>2<\/sup><br \/>\n(iii) sec<sup>4<\/sup>\u03b8\u2013 sec<sup>2<\/sup>\u03b8 = tan<sup>4<\/sup>\u03b8 + tan<sup>2<\/sup>\u03b8<br \/>\n<strong>Sol.\u00a0 \u00a0 (i)\u00a0<\/strong>We have,<br \/>\nLHS = (sin\u03b8 + cosec\u03b8)<sup>2<\/sup> + (cos\u03b8 + sec\u03b8)<sup>2<\/sup><br \/>\n= (sin<sup>2<\/sup>\u03b8 + cosec<sup>2<\/sup>\u03b8 + 2sin\u03b8 cosec\u03b8) (cos<sup>2<\/sup>\u03b8 + sec<sup>2<\/sup>\u03b8 + 2cos\u03b8 sec\u03b8)<br \/>\n\\(\\left( {{\\sin }^{2}}\\theta +\\cos e{{c}^{2}}\\theta +2\\sin \\theta .\\frac{1}{\\sin \\theta } \\right)+\\left( {{\\cos }^{2}}\\theta +{{\\sec }^{2}}\\theta +2\\cos \\theta .\\frac{1}{\\cos \\theta } \\right) \\)<br \/>\n= (sin<sup>2<\/sup>\u03b8 + cosec<sup>2<\/sup>\u03b8 + 2) + (cos<sup>2<\/sup>\u03b8 + sec<sup>2<\/sup>\u03b8 + 2)<br \/>\n= sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8 + cosec<sup>2<\/sup>\u03b8 + sec<sup>2<\/sup>\u03b8 + 4<br \/>\n= 1 + (1 + cot<sup>2<\/sup>\u03b8) + (1 + tan<sup>2<\/sup>\u03b8) + 4<br \/>\n[\u2235 cosec2\u03b8 = 1 + cot<sup>2<\/sup>\u03b8, sec<sup>2<\/sup>\u03b8 = 1 + tan<sup>2<\/sup>\u03b8]<br \/>\n= 7 + tan<sup>2<\/sup>\u03b8 + cot<sup>2<\/sup>\u03b8 = RHS.<br \/>\n<strong>(ii)<\/strong>\u00a0\u00a0We have,<br \/>\nLHS = (sin\u03b8 + sec\u03b8)<sup>2<\/sup> + (cos\u03b8 + cosec\u03b8)<sup>2<\/sup><br \/>\n\\( ={{\\left( \\sin \\theta +\\frac{1}{\\cos \\theta } \\right)}^{2}}+{{\\left( \\cos \\theta +\\frac{1}{\\sin \\theta } \\right)}^{2}} \\)<br \/>\n\\( ={{\\sin }^{2}}\\theta +\\frac{1}{{{\\cos }^{2}}\\theta }+\\frac{2\\sin \\theta }{\\cos \\theta }+{{\\cos }^{2}}\\theta +\\frac{1}{{{\\sin }^{2}}\\theta }+\\frac{2\\cos \\theta }{\\sin \\theta } \u00a0\\)<br \/>\n\\( =({{\\sin }^{2}}\\theta +{{\\cos }^{2}}\\theta )+\\left( \\frac{1}{{{\\cos }^{2}}\\theta }+\\frac{1}{{{\\sin }^{2}}\\theta } \\right)+2\\left( \\frac{\\sin \\theta }{\\cos \\theta }+\\frac{\\cos \\theta }{\\sin \\theta } \\right) \\)<br \/>\n\\( =({{\\sin }^{2}}\\theta +{{\\cos }^{2}}\\theta )+\\left( \\frac{{{\\sin }^{2}}\\theta +{{\\cos }^{2}}\\theta }{{{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta } \\right)+\\frac{2({{\\sin }^{2}}\\theta +{{\\cos }^{2}}\\theta )}{\\sin \\theta \\cos \\theta } \\)<br \/>\n\\( =1+\\frac{1}{{{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta }+\\frac{2}{\\sin \\theta \\cos \\theta } \u00a0\\)<br \/>\n\\( ={{\\left( 1+\\frac{1}{\\sin \\theta \\cos \\theta } \\right)}^{2}} \\)<br \/>\n= (1 + sec\u03b8 cosec\u03b8)<sup>2<\/sup>\u00a0= RHS<br \/>\n<strong>(iii)<\/strong>\u00a0\u00a0We have,<br \/>\nLHS = sec<sup>4<\/sup>\u03b8\u2013 sec<sup>2<\/sup>\u03b8<br \/>\n= sec<sup>2<\/sup>\u03b8 (sec<sup>2<\/sup>\u03b8 \u2013 1) = (1 + tan<sup>2<\/sup>\u03b8) (1 + tan<sup>2<\/sup>\u03b8 \u2013 1)<br \/>\n[ sec<sup>2<\/sup>\u03b8 = 1 + tan<sup>2\u00a0<\/sup>\u03b8]<br \/>\n= (1 + tan<sup>2<\/sup>\u03b8) tan<sup>2<\/sup>\u03b8<br \/>\n= tan<sup>4<\/sup>\u03b8 + tan<sup>2<\/sup>\u03b8\u00a0= RHS<\/p>\n<p><strong>Example 4:<\/strong> \u00a0 \u00a0Prove the following identities:<br \/>\n(i) cos4<sup>4<\/sup> A \u2013 cos<sup>2<\/sup> A = sin<sup>4<\/sup> A \u2013 sin<sup>2<\/sup> A<br \/>\n(ii) cot<sup>4<\/sup> A \u2013 1 = cosec<sup>4<\/sup> A \u2013 2cosec<sup>2<\/sup> A<br \/>\n(iii) sin<sup>6<\/sup> A + cos<sup>6<\/sup> A = 1 \u2013 3sin<sup>2<\/sup> A cos<sup>2<\/sup> A.<br \/>\n<strong>Sol.(i)<\/strong> We have,<br \/>\nLHS = cos4<sup>4<\/sup> A \u2013 cos<sup>2<\/sup> A = cos<sup>2<\/sup>A (cos<sup>2<\/sup>A \u2013 1)<br \/>\n= \u2013 cos<sup>2<\/sup> A (1 \u2013 cos<sup>2<\/sup> A) = \u2013 cos<sup>2<\/sup>A sin2A<br \/>\n= \u2013(1 \u2013 sin<sup>2<\/sup> A) sin<sup>2<\/sup> A = \u2013 sin<sup>2<\/sup> A + sin<sup>4<\/sup> A<br \/>\n= sin<sup>4<\/sup> A \u2013 sin<sup>2<\/sup> A = RHS<br \/>\n<strong>(ii)<\/strong> We have,<br \/>\nLHS = cot<sup>4<\/sup> A \u2013 1 = (cosec<sup>2<\/sup>A \u2013 1)<sup>2<\/sup> \u2013 1<br \/>\n[\u2235 cot<sup>2<\/sup>A = cosec<sup>2<\/sup>A \u20131 \u00a0 \u21d2\u00a0\u00a0\u00a0cot<sup>4<\/sup>A = (cosec<sup>2<\/sup>A \u2013 1)<sup>2<\/sup>]<br \/>\n= cosec<sup>4<\/sup>A \u2013 2 cosec<sup>2<\/sup>A + 1 \u2013 1<br \/>\n= cosec<sup>4<\/sup> A \u2013 2cosec<sup>2<\/sup>\u00a0A = RHS<br \/>\n<strong>(iii)<\/strong> We have,<br \/>\nLHS = sin<sup>6<\/sup> A + cos<sup>6<\/sup> A = (sin<sup>2<\/sup> A)<sup>3<\/sup> + (cos2 A)<sup>3<\/sup><br \/>\n= (sin<sup>2<\/sup> A + cos<sup>2<\/sup> A) {(sin<sup>2<\/sup> A)<sup>2<\/sup> + (cos<sup>2<\/sup> A)<sup>2<\/sup>\u00a0\u2013 sin<sup>2<\/sup> A cos<sup>2<\/sup> A)}<br \/>\n[\u2235 a<sup>3<\/sup> + b<sup>3<\/sup> = (a + b) (a<sup>2<\/sup> \u2013 ab + b<sup>2<\/sup>)]<br \/>\n={(sin<sup>2<\/sup> A)<sup>2<\/sup> + (cos<sup>2<\/sup> A)<sup>2<\/sup> + 2 sin<sup>2<\/sup> A cos<sup>2<\/sup> A \u2013 sin<sup>2<\/sup> A cos<sup>2<\/sup> A}<br \/>\n= [(sin<sup>2<\/sup> A + cos<sup>2<\/sup> A)<sup>2<\/sup> \u2013 3 sin<sup>2<\/sup> A cos<sup>2<\/sup> A]<br \/>\n= 1 \u2013 3sin<sup>2<\/sup> A cos<sup>2<\/sup> A = RHS<\/p>\n<p><strong>Example 5:<\/strong> \u00a0 \u00a0Prove the following identities:<br \/>\n\\( \\left( \\text{i} \\right)\\frac{si{{n}^{2}}A}{co{{s}^{2}}A}+\\frac{co{{s}^{2}}A}{si{{n}^{2}}A}=\\frac{1}{si{{n}^{2}}A\\,co{{s}^{2}}A}-2 \\)<br \/>\n\\( \\left( \\text{ii} \\right)\\frac{cosA}{1tanA}+\\frac{si{{n}^{2}}A}{sinAcosA}=\\sin A\\text{ }+\\cos A \\)<br \/>\n\\( \\left( \\text{iii} \\right)\\frac{{{(1+sin\\,\\theta )}^{2}}+{{(1sin\\,\\theta )}^{2}}}{co{{s}^{2}}\\theta }=2\\left( \\frac{1+si{{n}^{2}}\\,\\theta }{1-si{{n}^{2}}\\,\\theta } \\right) \\)<br \/>\n<strong>Sol. \u00a0 \u00a0(i)<\/strong> We have,<br \/>\n\\( LHS=\\frac{si{{n}^{2}}A}{co{{s}^{2}}A}+\\frac{co{{s}^{2}}A}{si{{n}^{2}}A}=\\frac{si{{n}^{4}}\\,A+co{{s}^{2}}A}{si{{n}^{2}}\\,Aco{{s}^{2}}A} \\)<br \/>\n\\( =\\frac{{{({{\\sin }^{2}}A)}^{2}}+{{({{\\cos }^{2}}A)}^{2}}+2{{\\sin }^{2}}A{{\\cos }^{2}}A-2{{\\sin }^{2}}A{{\\cos }^{2}}A}{{{\\sin }^{2}}A{{\\cos }^{2}}A} \\)<br \/>\n\\( =\\frac{{{({{\\sin }^{2}}A+{{\\cos }^{2}}A)}^{2}}-2{{\\sin }^{2}}A{{\\cos }^{2}}A}{{{\\sin }^{2}}A{{\\cos }^{2}}A} \\)<br \/>\n\\(=\\frac{1-2{{\\sin }^{2}}A{{\\cos }^{2}}A}{{{\\sin }^{2}}A{{\\cos }^{2}}A} \\)<br \/>\n\\(=\\frac{1}{{{\\sin }^{2}}A{{\\cos }^{2}}A}-2=RHS\\)<br \/>\n<strong>(ii)<\/strong> We have,<br \/>\n\\( LHS=\\frac{\\cos A}{1-\\tan A}+\\frac{{{\\sin }^{2}}A}{\\sin A-\\cos A} \\)<br \/>\n\\( =\\frac{\\cos A}{1-\\frac{\\sin A}{\\cos A}}+\\frac{{{\\sin }^{2}}A}{\\sin A-\\cos A} \\)<br \/>\n\\( =\\frac{\\cos A}{\\frac{\\cos A-\\sin A}{\\cos A}}+\\frac{{{\\sin }^{2}}A}{\\sin A-\\cos A} \\)<br \/>\n\\( =\\frac{{{\\cos }^{2}}A}{\\cos A\\sin A}+\\frac{{{\\sin }^{2}}A}{\\sin A\\cos A} \\)<br \/>\n\\( =\\frac{{{\\cos }^{2}}A}{\\cos A\\sin A}-\\frac{{{\\sin }^{2}}A}{\\cos A\\sin A} \\)<br \/>\n\\( =\\frac{{{\\cos }^{2}}A-{{\\sin }^{2}}A}{\\cos A-\\sin A} \\)<br \/>\n\\( =\\frac{(\\cos A+\\sin A)\\,(\\cos A-\\sin A)}{\\cos A-\\sin A} \\)<br \/>\n= cos A + sin A = RHS<br \/>\n<strong>(iii)<\/strong> We have,<br \/>\n\\( LHS=\\frac{{{(1+\\sin \\theta )}^{2}}+{{(1\\sin \\theta )}^{2}}}{{{\\cos }^{2}}\\theta } \\)<br \/>\n\\( =\\frac{(1+2\\sin \\theta +{{\\sin }^{2}}\\theta )+(12\\sin \\theta +{{\\sin }^{2}}\\theta )}{{{\\cos }^{2}}\\theta } \\)<br \/>\n\\( =\\frac{2+2{{\\sin }^{2}}\\theta }{{{\\cos }^{2}}\\theta }=\\frac{2\\,(1+{{\\sin }^{2}}\\theta )}{1-{{\\sin }^{2}}\\theta }=2\\left( \\frac{1+{{\\sin }^{2}}\\theta }{1-{{\\sin }^{2}}\\theta } \\right) \\)<br \/>\n= RHS.<\/p>\n<p><strong>Example 6:<\/strong> \u00a0 \u00a0Prove the following identities:<br \/>\n(i) 2 (sin<sup>6<\/sup> \u03b8 + cos<sup>6<\/sup> \u03b8) \u20133(sin<sup>4<\/sup> \u03b8 + cos<sup>4<\/sup> \u03b8) + 1 = 0<br \/>\n(ii) (sin<sup>8<\/sup> \u03b8 \u2013 cos<sup>8<\/sup>\u03b8) = (sin<sup>2<\/sup> \u03b8 \u2013 cos<sup>2<\/sup> \u03b8) (1 \u2013 2sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup>\u03b8)<br \/>\n<strong>Sol. \u00a0 \u00a0(i)<\/strong> We have,<br \/>\nLHS = 2 (sin<sup>6<\/sup> \u03b8 + cos<sup>6<\/sup> \u03b8) \u20133(sin<sup>4<\/sup> \u03b8 + cos<sup>4<\/sup> \u03b8) + 1<br \/>\n= 2 [(sin<sup>2<\/sup> \u03b8)<sup>3<\/sup> + (cos<sup>2<\/sup>\u03b8)<sup>3<\/sup>] \u2013 [3 (sin<sup>2<\/sup> \u03b8)<sup>2<\/sup> + (cos<sup>2<\/sup> \u03b8)<sup>2<\/sup>] + 1<br \/>\n= 2[(sin<sup>2<\/sup> \u03b8 + cos<sup>2<\/sup>\u03b8) {(sin<sup>2<\/sup>\u03b8)<sup>2<\/sup> + (cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> \u2013 sin<sup>2<\/sup>\u03b8 cos<sup>2<\/sup> \u03b8)]} \u2013 3[(sin<sup>2<\/sup> \u03b8)<sup>2<\/sup> + (cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> + 2 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8 \u20132 sin2 \u03b8 cos2 \u03b8] + 1<br \/>\n= 2[(sin<sup>2<\/sup> \u03b8)<sup>2<\/sup>\u00a0+ (cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> + 2 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8 \u20133 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8] \u20133 [(sin<sup>2<\/sup> \u03b8 + cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> \u2013 2 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8] + 1<br \/>\n= 2[(sin<sup>2<\/sup> \u03b8 + cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> \u2013 3 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8] \u20133 [1 \u2013 2 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8] + 1<br \/>\n= 2 (1 \u2013 3 sin<sup>2<\/sup>\u03b8 cos<sup>2<\/sup>\u03b8) \u2013 3(1 \u2013 2 sin<sup>2<\/sup>\u03b8 cos<sup>2<\/sup>\u03b8) + 1<br \/>\n= 2 \u2013 6 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup>\u03b8 \u20133 + 6 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8 + 1<br \/>\n= 0 = RHS<br \/>\n<strong>(ii)<\/strong> We have,<br \/>\nLHS = (sin<sup>8<\/sup> \u03b8 \u2013 cos<sup>8<\/sup>\u03b8) = (sin<sup>4<\/sup> \u03b8)<sup>2<\/sup> \u2013 (cos<sup>4<\/sup> \u03b8)<sup>2<\/sup><br \/>\n= (sin<sup>4<\/sup> \u03b8 \u2013 cos<sup>4<\/sup> \u03b8) (sin<sup>4<\/sup> \u03b8 + cos<sup>4<\/sup> \u03b8)<br \/>\n= (sin<sup>2<\/sup> \u03b8 \u2013 cos<sup>2<\/sup> \u03b8) (sin<sup>2<\/sup> \u03b8 + cos<sup>2<\/sup> \u03b8) (sin<sup>4<\/sup> \u03b8 + cos<sup>4<\/sup> \u03b8)<br \/>\n= (sin<sup>2<\/sup> \u03b8 \u2013 cos<sup>2<\/sup> \u03b8){(sin<sup>2<\/sup> \u03b8)<sup>2<\/sup> + (cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> + 2 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8 \u2013 2 sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8\uf07d<br \/>\n= (sin<sup>2<\/sup> \u03b8 \u2013 cos<sup>2<\/sup> \u03b8) {(sin<sup>2<\/sup> \u03b8 + cos<sup>2<\/sup> \u03b8)<sup>2<\/sup> \u2013 2sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup> \u03b8}<br \/>\n= (sin<sup>2<\/sup> \u03b8 \u2013 cos<sup>2<\/sup> \u03b8) (1 \u2013 2sin<sup>2<\/sup> \u03b8 cos<sup>2<\/sup>\u03b8) = RHS<\/p>\n<p><strong>Example 7:<\/strong>\u00a0 \u00a0 If (secA + tanA)(secB + tanB)(secC + tanC) = (secA \u2013 tanA)(secB \u2013 tanB)(secC \u2013 tanC) \u00a0prove that each of the side is equal to \u00b11.<br \/>\nWe have,<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0(secA + tanA)(secB + tanB)(secC + tanC)\u00a0= (secA \u2013 tanA)(secB \u2013 tanB)(secC \u2013 tanC)<br \/>\nMultiplying both sides by<br \/>\n(secA \u2013 tanA)(secB \u2013 tanB)(secC \u2013 tanC) we get<br \/>\n(secA + tanA) (secB + tanB) (secC + tanC) (secA \u2013 tanA) (secB \u2013 tanB) (secC \u2013 tanC)\u00a0= (secA \u2013 tanA)<sup>2<\/sup> (secB \u2013 tanB)<sup>2<\/sup> (secC \u2013 tanC)<sup>2<\/sup><br \/>\n(sec<sup>2<\/sup>A \u2013 tan<sup>2<\/sup>A)(sec<sup>2<\/sup>B \u2013 tan<sup>2<\/sup>B) (sec<sup>2<\/sup>C \u2013 tan<sup>2<\/sup>C)\u00a0= (secA \u2013 tanA)2(secB \u2013 tanB)2(secC \u2013 tanC)<sup>2<\/sup><br \/>\n1 = [(secA \u2013 tanA)(secB \u2013 tanB) (secC \u2013 tanC)]<sup>2<\/sup><br \/>\n(secA \u2013 tanA)(secB \u2013 tanB)(secC \u2013 tanC) = \u00b11<br \/>\nSimilarly, multiplying both sides by<br \/>\n(secA + tanA)(secB + tanB)(secC + tanC),<br \/>\nwe get<br \/>\n(secA + tanA)(secB + tanB)(secC + tanC) = \u00b11<\/p>\n<p><strong>Example 8:<\/strong>\u00a0 \u00a0 If tan\u03b8 + sin\u03b8 = m and tan\u03b8 \u2013 sin\u03b8 = n, show that m<sup>2<\/sup> \u2013 n<sup>2<\/sup> = \u00a0\\(4\\sqrt{mn}\\).<br \/>\n<strong>Sol. \u00a0<\/strong>\u00a0We have,<br \/>\nLHS = m<sup>2<\/sup> \u2013 n<sup>2<\/sup> = (tan\u03b8 + sin\u03b8)<sup>2<\/sup> \u2013 (tan\u03b8 \u2013 sin\u03b8)<sup>2<\/sup><br \/>\n= 4tan\u03b8 sin\u03b8 \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0(a + b)<sup>2<\/sup> \u2013 (a \u2013 b)<sup>2<\/sup> = 4ab]<br \/>\n\\( =4\\sqrt{(\\tan \\theta +\\sin \\theta )(\\tan \\theta \\sin \\theta )} \\)<br \/>\n\\( =4\\sqrt{{{\\tan }^{2}}\\theta {{\\sin }^{2}}\\theta } \\)<br \/>\n\\( =4\\sqrt{\\frac{{{\\sin }^{2}}\\theta }{{{\\cos }^{2}}\\theta }{{\\sin }^{2}}\\theta } \\\\ \\)<br \/>\n\\( =4\\sqrt{\\frac{{{\\sin }^{2}}\\theta {{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta }{{{\\cos }^{2}}\\theta }} \\)<br \/>\n\\( =4\\sqrt{\\frac{{{\\sin }^{2}}\\theta (1{{\\cos }^{2}}\\theta )}{{{\\cos }^{2}}\\theta }}=4\\sqrt{\\frac{{{\\sin }^{4}}\\theta }{{{\\cos }^{2}}\\theta }} \\)<br \/>\n\\( =\\text{ }4\\frac{{{\\sin }^{2}}\\theta }{\\cos \\theta }=4\\sin \\theta \\frac{\\sin \\theta }{\\cos \\theta }=4\\sin \\theta \\tan \\theta \\)<br \/>\nAnd, RHS = \\(4\\sqrt{mn}\\)<\/p>\n<p><strong>Example 9:<\/strong>\u00a0 \u00a0 If cos\u03b8 + sin\u03b8 = \u221a2 cos\u03b8, show that<br \/>\ncos\u03b8 \u2013 sin\u03b8 = \u221a2 sin\u03b8.<br \/>\n<strong>Sol. \u00a0<\/strong>\u00a0 We have,<br \/>\ncos\u03b8 + sin\u03b8 = cos\u03b8<br \/>\n\u21d2\u00a0(cos\u03b8 + sin\u03b8)2 = 2 cos<sup>2<\/sup>\u03b8<br \/>\n\u21d2\u00a0cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8 + 2 cos\u03b8sin\u03b8 = 2 cos<sup>2<\/sup>\u03b8<br \/>\n\u21d2\u00a0cos<sup>2<\/sup>\u03b8 \u2013 2cos\u03b8 sin\u03b8 = sin2\u03b8<br \/>\n\u21d2\u00a0cos<sup>2<\/sup>\u03b8 \u2013 2cos\u03b8sin\u03b8 + sin<sup>2<\/sup>\u03b8 = 2sin<sup>2<\/sup>\u03b8<br \/>\n\u21d2\u00a0(cos\u03b8 \u2013 sin\u03b8)<sup>2<\/sup>\u00a0= 2sin<sup>2<\/sup>\u03b8<br \/>\n\u21d2\u00a0cos\u03b8 \u2013 sin\u03b8 = \u221a2 sin\u03b8<\/p>\n<p><strong>Example 10:<\/strong>\u00a0 \u00a0 If sin\u03b8 + cos\u03b8 = p and sec\u03b8 + cosec\u03b8 = q, show that q(p<sup>2<\/sup> \u2013 1) = 2p<br \/>\n<strong>Sol. \u00a0<\/strong>\u00a0 We have,<br \/>\nLHS = q(p<sup>2<\/sup> \u2013 1)\u00a0= (sec\u03b8 + cosec\u03b8) [(sin\u03b8 + cos\u03b8)<sup>2<\/sup> \u2013 1]<br \/>\n\\( =\\left( \\frac{1}{\\cos \\theta }+\\frac{1}{\\sin \\theta } \\right)\\{\\sin 2\\theta +\\text{cos}2\\theta +2\\sin \\theta \\cos \\theta 1\\} \\)<br \/>\n\\( =\\left( \\frac{\\sin \\theta +\\cos \\theta }{\\cos \\theta \\sin \\theta } \\right)(1+2\\sin \\theta \\cos \\theta 1) \\)<br \/>\n\\( =\\left( \\frac{\\sin \\theta +\\cos \\theta }{\\cos \\theta \\sin \\theta } \\right)2\\sin \\theta \\cos \\ \\)<br \/>\n= 2(sin\u03b8 + cos\u03b8) = 2p = RHS<\/p>\n<p><strong>Example 11:<\/strong>\u00a0 \u00a0 If sec\u03b8 + tan\u03b8 = p, show that \u00a0 \\( \\frac{{{p}^{2}}-1}{{{p}^{2}}+1}=\\sin \\theta \\)<br \/>\n<strong>Sol. <\/strong>\u00a0 \u00a0 We have,<br \/>\n\\( =\\frac{{{\\sec }^{2}}\\theta +{{\\tan }^{2}}\\theta +2\\sec \\theta \\tan \\theta -1}{{{\\sec }^{2}}\\theta +{{\\tan }^{2}}\\theta +2\\sec \\theta \\tan \\theta +1} \\)<br \/>\n\\( =\\frac{({{\\sec }^{2}}\\theta -1)+{{\\tan }^{2}}\\theta +2\\sec \\theta \\tan \\theta }{{{\\sec }^{2}}\\theta +2\\sec \\theta \\tan \\theta +(1+{{\\tan }^{2}}\\theta )} \\)<br \/>\n\\( =\\frac{{{\\tan }^{2}}\\theta +{{\\tan }^{2}}\\theta +2\\sec \\theta \\tan \\theta }{{{\\sec }^{2}}\\theta +2\\sec \\theta \\tan \\theta +{{\\sec }^{2}}\\theta } \\)<br \/>\n\\( =\\frac{2{{\\tan }^{2}}\\theta +2\\tan \\theta \\sec \\theta }{2{{\\sec }^{2}}\\theta +2\\sec \\theta \\tan \\theta } \\)<br \/>\n\\( =\\frac{2\\tan \\theta \\,(\\tan \\theta +\\sec \\theta )}{2\\sec \\theta (\\sec \\theta +\\tan \\theta )} \\\\ \\)<br \/>\n\\( =\\frac{\\tan \\theta }{\\sec \\theta }=\\frac{\\sin \\theta }{\\cos \\theta \\sec \\theta } \\)<br \/>\n= sin\u03b8 = RHS<\/p>\n<p><strong>Example 12:<\/strong>\u00a0 \u00a0\u00a0\\(If\\frac{\\cos \\alpha }{\\cos \\beta }=m\\text{ and }\\frac{\\cos \\alpha }{\\sin \\beta }=n \\) \u00a0show that \u00a0 (m<sup>2<\/sup> + n<sup>2<\/sup>) cos<sup>2<\/sup>\u00a0\u03b2 = n<sup>2<\/sup>.<br \/>\n<strong>Sol. <\/strong>\u00a0 \u00a0LHS =\u00a0(m<sup>2<\/sup> + n<sup>2<\/sup>) cos<sup>2<\/sup>\u00a0\u03b2<br \/>\n\\( =\\left( \\frac{{{\\cos }^{2}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{2}}\\alpha }{{{\\sin }^{2}}\\beta } \\right)\\,{{\\cos }^{2}}\\beta \\text{ }\\left[ \\because \\ \\ m=\\frac{\\cos \\alpha }{\\cos \\beta }\\,\\,and\\,\\,n=\\frac{\\cos \\alpha }{\\sin \\beta } \\right] \\)<br \/>\n\\( =\\left( \\frac{{{\\cos }^{2}}\\alpha {{\\sin }^{2}}\\beta +{{\\cos }^{2}}\\alpha {{\\cos }^{2}}\\beta }{{{\\cos }^{2}}\\beta {{\\sin }^{2}}\\beta } \\right){{\\cos }^{2}}\\beta \\)<br \/>\n\\( ={{\\cos }^{2}}\\alpha \\left( \\frac{1}{{{\\cos }^{2}}\\beta {{\\sin }^{2}}\\beta } \\right){{\\cos }^{2}}\\beta \\\\ \\)<br \/>\n\\(=\\frac{{{\\cos }^{2}}\\alpha }{{{\\sin }^{2}}\\beta }={{\\left( \\frac{\\cos \\alpha }{\\sin \\beta } \\right)}^{2}} \\)<br \/>\n=\u00a0n<sup>2<\/sup> = RHS<\/p>\n<p><strong>Example 13:<\/strong>\u00a0 \u00a0 If acos\u03b8 + bsin\u03b8 = m and asin\u03b8 \u2013 bcos\u03b8 = n, prove that a<sup>2<\/sup> + b<sup>2<\/sup> = m<sup>2<\/sup> + n<sup>2<\/sup>.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 We have,<br \/>\nRHS = m<sup>2<\/sup> + n<sup>2<\/sup><br \/>\n= (acos\u03b8 + bsin\u03b8)<sup>2<\/sup> + (asin\u03b8 \u2013 bcos\u03b8)<sup>2<\/sup><br \/>\n= (a<sup>2<\/sup>cos2\u03b8 + b<sup>2<\/sup>sin2\u03b8 + 2ab cos\u03b8sin\u03b8) + (a<sup>2<\/sup> sin2\u03b8 + b<sup>2<\/sup>cos2\u03b8 \u2013 2ab sin\u03b8cos\u03b8)<br \/>\n= a<sup>2<\/sup>(cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8) + b<sup>2<\/sup>(sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8)<br \/>\n= a<sup>2<\/sup> + b<sup>2<\/sup> = LHS.<\/p>\n<p><strong>Example 14:<\/strong>\u00a0 \u00a0 If acos\u03b8 \u2013 bsin\u03b8 = c, prove that\u00a0asin\u03b8 + bcos\u03b8 = \\(\\pm \\sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \\)<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 We have,<br \/>\n(acos\u03b8 \u2013 bsin\u03b8)<sup>2<\/sup> + (asin\u03b8 + bcos\u03b8)<sup>2<\/sup><br \/>\n= (a<sup>2<\/sup>cos<sup>2<\/sup>\u03b8 + b<sup>2<\/sup>sin<sup>2<\/sup>\u03b8 \u2013 2ab sin\u03b8cos\u03b8) + (a<sup>2<\/sup>sin<sup>2<\/sup>\u03b8 + b<sup>2<\/sup>cos<sup>2<\/sup>\u03b8 + 2absin\u03b8cos\u03b8)<br \/>\n= a<sup>2<\/sup>(cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8) + b<sup>2<\/sup>(sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8)<br \/>\n= a<sup>2<\/sup> + b<sup>2<\/sup><br \/>\n\u21d2\u00a0c<sup>2<\/sup> + (asin\u03b8 + bcos\u03b8)<sup>2<\/sup> = a<sup>2<\/sup> + b<sup>2<\/sup>\u00a0 \u00a0 \u00a0 \u00a0 [\u2235\u00a0 acos\u03b8 \u2013 bsin\u03b8 = c]<br \/>\n\u21d2\u00a0(asin\u03b8 + bcos\u03b8)<sup>2<\/sup> = a<sup>2<\/sup> + b<sup>2<\/sup> \u2013 c<sup>2<\/sup><br \/>\n\u21d2\u00a0asin\u03b8 + bcos\u03b8 =\u00a0\\(\\pm \\sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \\)<\/p>\n<p><strong>Example 15:<\/strong>\u00a0 \u00a0 Prove that:<br \/>\n(1 \u2013 sin\u03b8 + cos\u03b8)<sup>2<\/sup> = 2(1 + cos\u03b8)(1 \u2013 sin\u03b8)<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 (1 \u2013 sin\u03b8 + cos\u03b8)<sup>2<\/sup><br \/>\n= 1 + sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8 \u2013 2sin\u03b8 + 2cos\u03b8 \u2013 2sin\u03b8cos\u03b8<br \/>\n= 2 \u2013 2sin\u03b8 + 2cos\u03b8 \u2013 2sin\u03b8cos\u03b8<br \/>\n= 2 (1 \u2013 sin\u03b8) + 2 cos\u03b8 (1 \u2013 sin\u03b8)<br \/>\n= 2(1 \u2013 sin\u03b8)(1 + cos\u03b8) = RHS<\/p>\n<p><strong>Example 16:<\/strong>\u00a0 \u00a0 If sin\u03b8 + sin<sup>2<\/sup>\u03b8 = 1, prove that cos<sup>2<\/sup>\u03b8 + cos<sup>4<\/sup>\u03b8 = 1.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 We have,<br \/>\nsin\u03b8 + sin<sup>2<\/sup>\u03b8 = 1<br \/>\n\u21d2\u00a0\u00a0sin\u03b8 = 1 \u2013 sin<sup>2<\/sup>\u03b8<br \/>\n\u21d2\u00a0\u00a0sin\u03b8 = cos<sup>2<\/sup>\u03b8<br \/>\nNow, cos<sup>2<\/sup>\u03b8 + cos<sup>4<\/sup>\u03b8\u00a0= cos<sup>2<\/sup>\u03b8 + (cos<sup>2<\/sup>\u03b8)<sup>2<\/sup><br \/>\n= cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8 = 1<\/p>\n<p><strong>Example 17:<\/strong>\u00a0 \u00a0\u00a0Prove that<br \/>\n\\( \\frac{\\sin \\theta -\\cos \\theta }{\\sin \\theta +\\cos \\theta }+\\frac{\\sin \\theta +\\cos \\theta }{\\sin \\theta -\\cos \\theta }=\\frac{2}{2{{\\sin }^{2}}\\theta -1} \\)<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \\( LHS=\\frac{\\sin \\theta -\\cos \\theta }{\\sin \\theta +\\cos \\theta }+\\frac{\\sin \\theta +\\cos \\theta }{\\sin \\theta -\\cos \\theta } \\)<br \/>\n\\( =\\frac{{{(\\sin \\theta -\\cos \\theta )}^{2}}+{{(\\sin \\theta +\\cos \\theta )}^{2}}}{(\\sin \\theta +\\cos \\theta )(\\sin \\theta -\\cos \\theta )} \\)<br \/>\n\\( =\\frac{2({{\\sin }^{2}}\\theta +{{\\cos }^{2}}\\theta )}{{{\\sin }^{2}}\\theta -{{\\cos }^{2}}\\theta } \\)<br \/>\n\\( =\\frac{2}{{{\\sin }^{2}}\\theta -(1-{{\\sin }^{2}}\\theta )} \\)<br \/>\n\\( =\\frac{2}{(2{{\\sin }^{2}}\\theta -1)}=RHS. \\)<\/p>\n<p><strong>Example 18:<\/strong>\u00a0 \u00a0 \u00a0Express the ratios cos A, tan A and sec A in terms of sin A.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Since cos2A + sin2A = 1, therefore,<br \/>\ncos2A = 1 \u2013 sin2A, i.e., cos A =\u00a0\\(\\pm \\sqrt{1-{{\\sin }^{2}}A} \\)<br \/>\nThis gives<br \/>\ncos A = \\(\\sqrt{1-{{\\sin }^{2}}A} \\) \u00a0 (Why ?)<br \/>\nHence,<br \/>\n\\( \\tan A=\\frac{\\sin A}{\\cos A}=\\frac{\\sin A}{\\sqrt{1-{{\\sin }^{2}}A}}\\text{ and} \\)<br \/>\n\\( \\sec A=\\frac{1}{\\cos A}=\\frac{1}{\\sqrt{1-{{\\sin }^{2}}A}} \\)<\/p>\n<p><strong>Example 19:<\/strong>\u00a0 \u00a0 \u00a0Prove that \u00a0 \\( \\frac{\\sin \\theta -\\cos \\theta +1}{\\sin \\theta +\\cos \\theta -1}=\\frac{1}{\\sec \\theta -\\tan \\theta } \\) \u00a0 \u00a0 \u00a0using the identity sec<sup>2<\/sup>\u03b8 = 1 + tan<sup>2<\/sup>\u03b8.<br \/>\n\\( LHS=\\frac{\\sin \\theta -\\cos \\theta +1}{\\sin \\theta +\\cos \\theta -1}=\\frac{\\tan \\theta -1+\\sec \\theta }{\\tan \\theta +1-\\sec \\theta } \\)<br \/>\n\\( =\\frac{(\\tan \\theta +\\sec \\theta )-1}{(\\tan \\theta -\\sec \\theta )+1} \\)<br \/>\n\\( =\\frac{\\{(\\tan \\theta +\\sec \\theta )-1\\}\\,(\\tan \\theta -\\sec \\theta )}{\\{(\\tan \\theta -\\sec \\theta )+1\\}\\,(\\tan \\theta -\\sec \\theta )} \\)<br \/>\n\\( =\\frac{({{\\tan }^{2}}\\theta -{{\\sec }^{2}}\\theta )-(\\tan \\theta -\\sec \\theta )}{\\{\\tan \\theta -\\sec \\theta +1\\}\\,(\\tan \\theta -\\sec \\theta )} \\)<br \/>\n\\( =\\frac{-1-\\tan \\theta +\\sec \\theta }{(\\tan \\theta -\\sec \\theta +1)\\,(\\tan \\theta -\\sec \\theta )} \\)<br \/>\n\\( =\\frac{-1}{\\tan \\theta -\\sec \\theta }=\\frac{1}{\\sec \\theta -\\tan \\theta } \\)<br \/>\nwhich is the RHS of the identity, we are required to prove.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Trigonometric Identities (1) Conditional trigonometrical identities We have certain trigonometric identities. Like sin2 \u03b8 + cos2 \u03b8 = 1 and 1 + tan2 \u03b8 = sec2 \u03b8 etc. Such identities are identities in the sense that they hold for all value of the angles which satisfy the given condition among them and they are called [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[166,167,146],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Trigonometric Identities - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/trigonometric-identities\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Trigonometric Identities\" \/>\n<meta property=\"og:description\" content=\"Trigonometric Identities (1) Conditional trigonometrical identities We have certain trigonometric identities. 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