Trigonometric Identities
(1) \(\tan \theta =\frac{\sin \theta }{\cos \theta }\text{ (linear)}\)
Conditional trigonometrical identities
We have certain trigonometric identities.
Like sin2 θ + cos2 θ = 1 and 1 + tan2 θ = sec2 θ 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 conditional identities.
Trigonometric Identities With Examples
Example 1: Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Sol. (i) We have,
LHS = (1 – sin2θ) sec2θ
= cos2θ sec2θ [∵ 1 – sin2θ = cos2θ]
\( ={{\cos }^{2}}\theta \left( \frac{1}{{{\cos }^{2}}\theta } \right)\left[ \because \ \ \sec \theta =\frac{1}{\cos \theta } \right]\)
= 1 = RHS
(ii) We have,
LHS = cos2θ (1 + tan2θ)
= cos2θ sec2θ [∵ 1 + tan2θ = sec2θ]
\( ={{\cos }^{2}}\theta \left( \frac{1}{{{\cos }^{2}}\theta } \right)\left[ \because \ \ \sec \theta =\frac{1}{\cos \theta } \right]\)
= 1 = RHS
Example 2: Prove the following trigonometric identities:
\( (\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta +\cot \theta \)
\( (\text{ii})\text{ }\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }=\frac{\sec \theta +1}{\sec \theta -1} \)
Sol. (i) We have,
\( LHS=\frac{\sin \theta }{(1-\cos \theta )}\times \frac{(1+\cos \theta )}{(1+\cos \theta )} \)
[Multiplying numerator and denominator by (1 + cosθ)]
\( =\frac{sin\theta (1+cos\theta )}{1co{{s}^{2}}\theta }=\frac{\sin \theta (1+\cos \theta )}{{{\sin }^{2}}\theta } \)
[∵ 1 – cos2θ = sin2θ]
\( =\frac{1+\cos \theta }{\sin \theta }=\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta } \)
= cosecθ + cotθ = RHS
\( \left[ \because \ \ \frac{1}{\sin \theta }=\cos ec\theta \,\,and\,\frac{\cos \theta }{\sin \theta }=\cot \theta \right] \)
(ii) We have,
\( LHS=\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta } \)
\( \frac{\frac{\sin \theta }{\cos \theta }+\sin \theta }{\frac{\sin \theta }{\cos \theta }-\sin \theta }=\frac{\sin \theta \left( \frac{1}{\cos \theta }+1 \right)}{\sin \theta \left( \frac{1}{\cos \theta }-1 \right)} \)
\( \frac{\frac{1}{\cos \theta }+1}{\frac{1}{\cos \theta }-1}=\frac{\sec \theta +1}{\sec \theta -1}=RHS \)
Example 3: Prove the following identities:
(i) (sinθ + cosecθ)2 + (cosθ + secθ)2 = 7 + tan2θ + cot2θ
(ii) (sinθ + secθ)2 + (cosθ + cosecθ)2 = (1 + secθ cosecθ)2
(iii) sec4θ– sec2θ = tan4θ + tan2θ
Sol. (i) We have,
LHS = (sinθ + cosecθ)2 + (cosθ + secθ)2
= (sin2θ + cosec2θ + 2sinθ cosecθ) (cos2θ + sec2θ + 2cosθ secθ)
\(\left( {{\sin }^{2}}\theta +\cos e{{c}^{2}}\theta +2\sin \theta .\frac{1}{\sin \theta } \right)+\left( {{\cos }^{2}}\theta +{{\sec }^{2}}\theta +2\cos \theta .\frac{1}{\cos \theta } \right) \)
= (sin2θ + cosec2θ + 2) + (cos2θ + sec2θ + 2)
= sin2θ + cos2θ + cosec2θ + sec2θ + 4
= 1 + (1 + cot2θ) + (1 + tan2θ) + 4
[∵ cosec2θ = 1 + cot2θ, sec2θ = 1 + tan2θ]
= 7 + tan2θ + cot2θ = RHS.
(ii) We have,
LHS = (sinθ + secθ)2 + (cosθ + cosecθ)2
\( ={{\left( \sin \theta +\frac{1}{\cos \theta } \right)}^{2}}+{{\left( \cos \theta +\frac{1}{\sin \theta } \right)}^{2}} \)
\( ={{\sin }^{2}}\theta +\frac{1}{{{\cos }^{2}}\theta }+\frac{2\sin \theta }{\cos \theta }+{{\cos }^{2}}\theta +\frac{1}{{{\sin }^{2}}\theta }+\frac{2\cos \theta }{\sin \theta } \)
\( =({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )+\left( \frac{1}{{{\cos }^{2}}\theta }+\frac{1}{{{\sin }^{2}}\theta } \right)+2\left( \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \right) \)
\( =({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )+\left( \frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta } \right)+\frac{2({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}{\sin \theta \cos \theta } \)
\( =1+\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }+\frac{2}{\sin \theta \cos \theta } \)
\( ={{\left( 1+\frac{1}{\sin \theta \cos \theta } \right)}^{2}} \)
= (1 + secθ cosecθ)2 = RHS
(iii) We have,
LHS = sec4θ– sec2θ
= sec2θ (sec2θ – 1) = (1 + tan2θ) (1 + tan2θ – 1)
[ sec2θ = 1 + tan2 θ]
= (1 + tan2θ) tan2θ
= tan4θ + tan2θ = RHS
Example 4: Prove the following identities:
(i) cos44 A – cos2 A = sin4 A – sin2 A
(ii) cot4 A – 1 = cosec4 A – 2cosec2 A
(iii) sin6 A + cos6 A = 1 – 3sin2 A cos2 A.
Sol.(i) We have,
LHS = cos44 A – cos2 A = cos2A (cos2A – 1)
= – cos2 A (1 – cos2 A) = – cos2A sin2A
= –(1 – sin2 A) sin2 A = – sin2 A + sin4 A
= sin4 A – sin2 A = RHS
(ii) We have,
LHS = cot4 A – 1 = (cosec2A – 1)2 – 1
[∵ cot2A = cosec2A –1 ⇒ cot4A = (cosec2A – 1)2]
= cosec4A – 2 cosec2A + 1 – 1
= cosec4 A – 2cosec2 A = RHS
(iii) We have,
LHS = sin6 A + cos6 A = (sin2 A)3 + (cos2 A)3
= (sin2 A + cos2 A) {(sin2 A)2 + (cos2 A)2 – sin2 A cos2 A)}
[∵ a3 + b3 = (a + b) (a2 – ab + b2)]
={(sin2 A)2 + (cos2 A)2 + 2 sin2 A cos2 A – sin2 A cos2 A}
= [(sin2 A + cos2 A)2 – 3 sin2 A cos2 A]
= 1 – 3sin2 A cos2 A = RHS
Example 5: Prove the following identities:
\( \left( \text{i} \right)\frac{si{{n}^{2}}A}{co{{s}^{2}}A}+\frac{co{{s}^{2}}A}{si{{n}^{2}}A}=\frac{1}{si{{n}^{2}}A\,co{{s}^{2}}A}-2 \)
\( \left( \text{ii} \right)\frac{cosA}{1tanA}+\frac{si{{n}^{2}}A}{sinAcosA}=\sin A\text{ }+\cos A \)
\( \left( \text{iii} \right)\frac{{{(1+sin\,\theta )}^{2}}+{{(1sin\,\theta )}^{2}}}{co{{s}^{2}}\theta }=2\left( \frac{1+si{{n}^{2}}\,\theta }{1-si{{n}^{2}}\,\theta } \right) \)
Sol. (i) We have,
\( LHS=\frac{si{{n}^{2}}A}{co{{s}^{2}}A}+\frac{co{{s}^{2}}A}{si{{n}^{2}}A}=\frac{si{{n}^{4}}\,A+co{{s}^{2}}A}{si{{n}^{2}}\,Aco{{s}^{2}}A} \)
\( =\frac{{{({{\sin }^{2}}A)}^{2}}+{{({{\cos }^{2}}A)}^{2}}+2{{\sin }^{2}}A{{\cos }^{2}}A-2{{\sin }^{2}}A{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \)
\( =\frac{{{({{\sin }^{2}}A+{{\cos }^{2}}A)}^{2}}-2{{\sin }^{2}}A{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \)
\(=\frac{1-2{{\sin }^{2}}A{{\cos }^{2}}A}{{{\sin }^{2}}A{{\cos }^{2}}A} \)
\(=\frac{1}{{{\sin }^{2}}A{{\cos }^{2}}A}-2=RHS\)
(ii) We have,
\( LHS=\frac{\cos A}{1-\tan A}+\frac{{{\sin }^{2}}A}{\sin A-\cos A} \)
\( =\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{{{\sin }^{2}}A}{\sin A-\cos A} \)
\( =\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{{{\sin }^{2}}A}{\sin A-\cos A} \)
\( =\frac{{{\cos }^{2}}A}{\cos A\sin A}+\frac{{{\sin }^{2}}A}{\sin A\cos A} \)
\( =\frac{{{\cos }^{2}}A}{\cos A\sin A}-\frac{{{\sin }^{2}}A}{\cos A\sin A} \)
\( =\frac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\cos A-\sin A} \)
\( =\frac{(\cos A+\sin A)\,(\cos A-\sin A)}{\cos A-\sin A} \)
= cos A + sin A = RHS
(iii) We have,
\( LHS=\frac{{{(1+\sin \theta )}^{2}}+{{(1\sin \theta )}^{2}}}{{{\cos }^{2}}\theta } \)
\( =\frac{(1+2\sin \theta +{{\sin }^{2}}\theta )+(12\sin \theta +{{\sin }^{2}}\theta )}{{{\cos }^{2}}\theta } \)
\( =\frac{2+2{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }=\frac{2\,(1+{{\sin }^{2}}\theta )}{1-{{\sin }^{2}}\theta }=2\left( \frac{1+{{\sin }^{2}}\theta }{1-{{\sin }^{2}}\theta } \right) \)
= RHS.
Example 6: Prove the following identities:
(i) 2 (sin6 θ + cos6 θ) –3(sin4 θ + cos4 θ) + 1 = 0
(ii) (sin8 θ – cos8θ) = (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2θ)
Sol. (i) We have,
LHS = 2 (sin6 θ + cos6 θ) –3(sin4 θ + cos4 θ) + 1
= 2 [(sin2 θ)3 + (cos2θ)3] – [3 (sin2 θ)2 + (cos2 θ)2] + 1
= 2[(sin2 θ + cos2θ) {(sin2θ)2 + (cos2 θ)2 – sin2θ cos2 θ)]} – 3[(sin2 θ)2 + (cos2 θ)2 + 2 sin2 θ cos2 θ –2 sin2 θ cos2 θ] + 1
= 2[(sin2 θ)2 + (cos2 θ)2 + 2 sin2 θ cos2 θ –3 sin2 θ cos2 θ] –3 [(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ] + 1
= 2[(sin2 θ + cos2 θ)2 – 3 sin2 θ cos2 θ] –3 [1 – 2 sin2 θ cos2 θ] + 1
= 2 (1 – 3 sin2θ cos2θ) – 3(1 – 2 sin2θ cos2θ) + 1
= 2 – 6 sin2 θ cos2θ –3 + 6 sin2 θ cos2 θ + 1
= 0 = RHS
(ii) We have,
LHS = (sin8 θ – cos8θ) = (sin4 θ)2 – (cos4 θ)2
= (sin4 θ – cos4 θ) (sin4 θ + cos4 θ)
= (sin2 θ – cos2 θ) (sin2 θ + cos2 θ) (sin4 θ + cos4 θ)
= (sin2 θ – cos2 θ){(sin2 θ)2 + (cos2 θ)2 + 2 sin2 θ cos2 θ – 2 sin2 θ cos2 θ
= (sin2 θ – cos2 θ) {(sin2 θ + cos2 θ)2 – 2sin2 θ cos2 θ}
= (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2θ) = RHS
Example 7: If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1.
We have,
Sol. (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC)
Multiplying both sides by
(secA – tanA)(secB – tanB)(secC – tanC) we get
(secA + tanA) (secB + tanB) (secC + tanC) (secA – tanA) (secB – tanB) (secC – tanC) = (secA – tanA)2 (secB – tanB)2 (secC – tanC)2
(sec2A – tan2A)(sec2B – tan2B) (sec2C – tan2C) = (secA – tanA)2(secB – tanB)2(secC – tanC)2
1 = [(secA – tanA)(secB – tanB) (secC – tanC)]2
(secA – tanA)(secB – tanB)(secC – tanC) = ±1
Similarly, multiplying both sides by
(secA + tanA)(secB + tanB)(secC + tanC),
we get
(secA + tanA)(secB + tanB)(secC + tanC) = ±1
Example 8: If tanθ + sinθ = m and tanθ – sinθ = n, show that m2 – n2 = \(4\sqrt{mn}\).
Sol. We have,
LHS = m2 – n2 = (tanθ + sinθ)2 – (tanθ – sinθ)2
= 4tanθ sinθ [∵ (a + b)2 – (a – b)2 = 4ab]
\( =4\sqrt{(\tan \theta +\sin \theta )(\tan \theta \sin \theta )} \)
\( =4\sqrt{{{\tan }^{2}}\theta {{\sin }^{2}}\theta } \)
\( =4\sqrt{\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }{{\sin }^{2}}\theta } \\ \)
\( =4\sqrt{\frac{{{\sin }^{2}}\theta {{\sin }^{2}}\theta {{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }} \)
\( =4\sqrt{\frac{{{\sin }^{2}}\theta (1{{\cos }^{2}}\theta )}{{{\cos }^{2}}\theta }}=4\sqrt{\frac{{{\sin }^{4}}\theta }{{{\cos }^{2}}\theta }} \)
\( =\text{ }4\frac{{{\sin }^{2}}\theta }{\cos \theta }=4\sin \theta \frac{\sin \theta }{\cos \theta }=4\sin \theta \tan \theta \)
And, RHS = \(4\sqrt{mn}\)
Example 9: If cosθ + sinθ = √2 cosθ, show that
cosθ – sinθ = √2 sinθ.
Sol. We have,
cosθ + sinθ = cosθ
⇒ (cosθ + sinθ)2 = 2 cos2θ
⇒ cos2θ + sin2θ + 2 cosθsinθ = 2 cos2θ
⇒ cos2θ – 2cosθ sinθ = sin2θ
⇒ cos2θ – 2cosθsinθ + sin2θ = 2sin2θ
⇒ (cosθ – sinθ)2 = 2sin2θ
⇒ cosθ – sinθ = √2 sinθ
Example 10: If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 – 1) = 2p
Sol. We have,
LHS = q(p2 – 1) = (secθ + cosecθ) [(sinθ + cosθ)2 – 1]
\( =\left( \frac{1}{\cos \theta }+\frac{1}{\sin \theta } \right)\{\sin 2\theta +\text{cos}2\theta +2\sin \theta \cos \theta 1\} \)
\( =\left( \frac{\sin \theta +\cos \theta }{\cos \theta \sin \theta } \right)(1+2\sin \theta \cos \theta 1) \)
\( =\left( \frac{\sin \theta +\cos \theta }{\cos \theta \sin \theta } \right)2\sin \theta \cos \ \)
= 2(sinθ + cosθ) = 2p = RHS
Example 11: If secθ + tanθ = p, show that \( \frac{{{p}^{2}}-1}{{{p}^{2}}+1}=\sin \theta \)
Sol. We have,
\( =\frac{{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta -1}{{{\sec }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta +1} \)
\( =\frac{({{\sec }^{2}}\theta -1)+{{\tan }^{2}}\theta +2\sec \theta \tan \theta }{{{\sec }^{2}}\theta +2\sec \theta \tan \theta +(1+{{\tan }^{2}}\theta )} \)
\( =\frac{{{\tan }^{2}}\theta +{{\tan }^{2}}\theta +2\sec \theta \tan \theta }{{{\sec }^{2}}\theta +2\sec \theta \tan \theta +{{\sec }^{2}}\theta } \)
\( =\frac{2{{\tan }^{2}}\theta +2\tan \theta \sec \theta }{2{{\sec }^{2}}\theta +2\sec \theta \tan \theta } \)
\( =\frac{2\tan \theta \,(\tan \theta +\sec \theta )}{2\sec \theta (\sec \theta +\tan \theta )} \\ \)
\( =\frac{\tan \theta }{\sec \theta }=\frac{\sin \theta }{\cos \theta \sec \theta } \)
= sinθ = RHS
Example 12: \(If\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n \) show that (m2 + n2) cos2 β = n2.
Sol. LHS = (m2 + n2) cos2 β
\( =\left( \frac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }+\frac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta } \right)\,{{\cos }^{2}}\beta \text{ }\left[ \because \ \ m=\frac{\cos \alpha }{\cos \beta }\,\,and\,\,n=\frac{\cos \alpha }{\sin \beta } \right] \)
\( =\left( \frac{{{\cos }^{2}}\alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta }{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \)
\( ={{\cos }^{2}}\alpha \left( \frac{1}{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \\ \)
\(=\frac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }={{\left( \frac{\cos \alpha }{\sin \beta } \right)}^{2}} \)
= n2 = RHS
Example 13: If acosθ + bsinθ = m and asinθ – bcosθ = n, prove that a2 + b2 = m2 + n2.
Sol. We have,
RHS = m2 + n2
= (acosθ + bsinθ)2 + (asinθ – bcosθ)2
= (a2cos2θ + b2sin2θ + 2ab cosθsinθ) + (a2 sin2θ + b2cos2θ – 2ab sinθcosθ)
= a2(cos2θ + sin2θ) + b2(sin2θ + cos2θ)
= a2 + b2 = LHS.
Example 14: If acosθ – bsinθ = c, prove that asinθ + bcosθ = \(\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \)
Sol. We have,
(acosθ – bsinθ)2 + (asinθ + bcosθ)2
= (a2cos2θ + b2sin2θ – 2ab sinθcosθ) + (a2sin2θ + b2cos2θ + 2absinθcosθ)
= a2(cos2θ + sin2θ) + b2(sin2θ + cos2θ)
= a2 + b2
⇒ c2 + (asinθ + bcosθ)2 = a2 + b2 [∵ acosθ – bsinθ = c]
⇒ (asinθ + bcosθ)2 = a2 + b2 – c2
⇒ asinθ + bcosθ = \(\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}} \)
Example 15: Prove that:
(1 – sinθ + cosθ)2 = 2(1 + cosθ)(1 – sinθ)
Sol. (1 – sinθ + cosθ)2
= 1 + sin2θ + cos2θ – 2sinθ + 2cosθ – 2sinθcosθ
= 2 – 2sinθ + 2cosθ – 2sinθcosθ
= 2 (1 – sinθ) + 2 cosθ (1 – sinθ)
= 2(1 – sinθ)(1 + cosθ) = RHS
Example 16: If sinθ + sin2θ = 1, prove that cos2θ + cos4θ = 1.
Sol. We have,
sinθ + sin2θ = 1
⇒ sinθ = 1 – sin2θ
⇒ sinθ = cos2θ
Now, cos2θ + cos4θ = cos2θ + (cos2θ)2
= cos2θ + sin2θ = 1
Example 17: Prove that
\( \frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2{{\sin }^{2}}\theta -1} \)
Sol. \( LHS=\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin \theta +\cos \theta }{\sin \theta -\cos \theta } \)
\( =\frac{{{(\sin \theta -\cos \theta )}^{2}}+{{(\sin \theta +\cos \theta )}^{2}}}{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )} \)
\( =\frac{2({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )}{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta } \)
\( =\frac{2}{{{\sin }^{2}}\theta -(1-{{\sin }^{2}}\theta )} \)
\( =\frac{2}{(2{{\sin }^{2}}\theta -1)}=RHS. \)
Example 18: Express the ratios cos A, tan A and sec A in terms of sin A.
Sol. Since cos2A + sin2A = 1, therefore,
cos2A = 1 – sin2A, i.e., cos A = \(\pm \sqrt{1-{{\sin }^{2}}A} \)
This gives
cos A = \(\sqrt{1-{{\sin }^{2}}A} \) (Why ?)
Hence,
\( \tan A=\frac{\sin A}{\cos A}=\frac{\sin A}{\sqrt{1-{{\sin }^{2}}A}}\text{ and} \)
\( \sec A=\frac{1}{\cos A}=\frac{1}{\sqrt{1-{{\sin }^{2}}A}} \)
Example 19: Prove that \( \frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta } \) using the identity sec2θ = 1 + tan2θ.
\( LHS=\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}=\frac{\tan \theta -1+\sec \theta }{\tan \theta +1-\sec \theta } \)
\( =\frac{(\tan \theta +\sec \theta )-1}{(\tan \theta -\sec \theta )+1} \)
\( =\frac{\{(\tan \theta +\sec \theta )-1\}\,(\tan \theta -\sec \theta )}{\{(\tan \theta -\sec \theta )+1\}\,(\tan \theta -\sec \theta )} \)
\( =\frac{({{\tan }^{2}}\theta -{{\sec }^{2}}\theta )-(\tan \theta -\sec \theta )}{\{\tan \theta -\sec \theta +1\}\,(\tan \theta -\sec \theta )} \)
\( =\frac{-1-\tan \theta +\sec \theta }{(\tan \theta -\sec \theta +1)\,(\tan \theta -\sec \theta )} \)
\( =\frac{-1}{\tan \theta -\sec \theta }=\frac{1}{\sec \theta -\tan \theta } \)
which is the RHS of the identity, we are required to prove.