Trigonometric Identities

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.