• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • ICSE Solutions
    • ICSE Solutions for Class 10
    • ICSE Solutions for Class 9
    • ICSE Solutions for Class 8
    • ICSE Solutions for Class 7
    • ICSE Solutions for Class 6
  • Selina Solutions
  • ML Aggarwal Solutions
  • ISC & ICSE Papers
    • ICSE Previous Year Question Papers Class 10
    • ISC Previous Year Question Papers
    • ICSE Specimen Paper 2021-2022 Class 10 Solved
    • ICSE Specimen Papers 2020 for Class 9
    • ISC Specimen Papers 2020 for Class 12
    • ISC Specimen Papers 2020 for Class 11
    • ICSE Time Table 2020 Class 10
    • ISC Time Table 2020 Class 12
  • Maths
    • Merit Batch

A Plus Topper

Improve your Grades

  • CBSE Sample Papers
  • HSSLive
    • HSSLive Plus Two
    • HSSLive Plus One
    • Kerala SSLC
  • Exams
  • NCERT Solutions for Class 10 Maths
  • NIOS
  • Chemistry
  • Physics
  • ICSE Books

Trigonometric Identities

April 30, 2023 by Veerendra

Trigonometric Identities

(1) \(\tan \theta =\frac{\sin \theta }{\cos \theta }\text{ (linear)}\)
Trigonometric Identities 1
Trigonometric Identities 2

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 3
Trigonometric Identities 4
Trigonometric Identities 5

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.

Filed Under: Mathematics Tagged With: Trigonometric Identities, Trigonometric Identities Examples, Trigonometry

Primary Sidebar

  • MCQ Questions
  • ICSE Solutions
  • Selina ICSE Solutions
  • Concise Mathematics Class 10 ICSE Solutions
  • Concise Physics Class 10 ICSE Solutions
  • Concise Chemistry Class 10 ICSE Solutions
  • Concise Biology Class 10 ICSE Solutions
  • Concise Mathematics Class 9 ICSE Solutions
  • Concise Physics Class 9 ICSE Solutions
  • Concise Chemistry Class 9 ICSE Solutions
  • Concise Biology Class 9 ICSE Solutions
  • ML Aggarwal Solutions
  • ML Aggarwal Class 10 Solutions
  • ML Aggarwal Class 9 Solutions
  • ML Aggarwal Class 8 Solutions
  • ML Aggarwal Class 7 Solutions
  • ML Aggarwal Class 6 Solutions
  • HSSLive Plus One
  • HSSLive Plus Two
  • Kerala SSLC
  • Recent Posts

    • The River Question and Answers
    • What is Man without the Beasts? Question and Answers
    • Not Just a Teacher, but a Friend Question and Answers
    • Swami is Expelled from School Question and Answers
    • V.V.S. Laxman, Very Very Special Question and Answers
    • What is a Player? Question and Answers
    • True Height Question and Answers
    • Little Bobby Question and Answers
    • The Duck and the Kangaroo Question and Answers
    • The Snake and the Mirror Question and Answers
    • AP Board Question and Answers 9th Class English

    Footer

    • Picture Dictionary
    • English Speech
    • ICSE Solutions
    • Selina ICSE Solutions
    • ML Aggarwal Solutions
    • HSSLive Plus One
    • HSSLive Plus Two
    • Kerala SSLC
    • Distance Education
    DisclaimerPrivacy Policy
    Area Volume Calculator