Introduction to Trigonometry | Exercise 8.3

Question 4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ(\text{cosec } \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}

(ii) cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A

(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ cosec θ\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta [Hint : Write the expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta]

(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A} [Hint : Simplify LHS and RHS separately]

(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cosec A+cot⁡A\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A, using the identity cosec2A=1+cot⁡2A\text{cosec}^2 A = 1 + \cot^2 A.

(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A

(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta

(viii) (sin⁡A+cosec A)2+(cos⁡A+sec⁡A)2=7+tan⁡2A+cot⁡2A(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A

(ix) (cosec A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A(\text{cosec } A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A} [Hint : Simplify LHS and RHS separately]

(x) (1+tan⁡2A1+cot⁡2A)=(1−tan⁡A1−cot⁡A)2=tan⁡2A\left(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A

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Solution
Understand the Question

To prove trigonometric identities involving acute angles:

  • Standard approach is to simplify the more complex side (usually LHS\text{LHS}) to match the other side (RHS\text{RHS}), or simplify both sides separately until they yield the same expression.
  • Key standard identities and relations used throughout:
    • Reciprocal Relations: cosec θ=1sin⁡θ\text{cosec } \theta = \dfrac{1}{\sin \theta}, sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{\cos \theta}, cot⁡θ=1tan⁡θ=cos⁡θsin⁡θ\cot \theta = \dfrac{1}{\tan \theta} = \dfrac{\cos \theta}{\sin \theta}, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{\sin \theta}{\cos \theta}
    • Pythagorean Identities: sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta, 1+cot⁡2θ=cosec2θ1 + \cot^2 \theta = \text{cosec}^2 \theta

(i) Prove (cosec θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ(\text{cosec } \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}

Step 1 · Simplify LHS using sin⁡θ\sin \theta and cos⁡θ\cos \theta

Converting into sin⁡θ\sin \theta and cos⁡θ\cos \theta:

LHS=(cosec θ−cot⁡θ)2=(1sin⁡θ−cos⁡θsin⁡θ)2=(1−cos⁡θsin⁡θ)2=(1−cos⁡θ)2sin⁡2θ\begin{aligned} \text{LHS} &= (\text{cosec } \theta - \cot \theta)^2 \\[0.6em] &= \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2 \\[0.6em] &= \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 \\[0.6em] &= \frac{(1 - \cos \theta)^2}{\sin^2 \theta} \end{aligned}

Using sin⁡2θ=1−cos⁡2θ=(1−cos⁡θ)(1+cos⁡θ)\sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta):

=(1−cos⁡θ)21−cos⁡2θ=(1−cos⁡θ)(1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)=1−cos⁡θ1+cos⁡θ=RHS\begin{aligned} &= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} \\[0.6em] &= \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} \\[0.6em] &= \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS} \end{aligned}
Answer

(i) LHS=RHS\text{LHS} = \text{RHS}

(ii) Prove cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A

Step 1 · Take common denominator and simplify LHS

Taking common denominator:

LHS=cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=cos⁡2A+(1+sin⁡A)2(1+sin⁡A)cos⁡A=cos⁡2A+1+2sin⁡A+sin⁡2A(1+sin⁡A)cos⁡A\begin{aligned} \text{LHS} &= \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} \\[0.6em] &= \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A) \cos A} \\[0.6em] &= \frac{\cos^2 A + 1 + 2 \sin A + \sin^2 A}{(1 + \sin A) \cos A} \end{aligned}

Using cos⁡2A+sin⁡2A=1\cos^2 A + \sin^2 A = 1:

=(cos⁡2A+sin⁡2A)+1+2sin⁡A(1+sin⁡A)cos⁡A=1+1+2sin⁡A(1+sin⁡A)cos⁡A=2+2sin⁡A(1+sin⁡A)cos⁡A=2(1+sin⁡A)(1+sin⁡A)cos⁡A=2cos⁡A=2sec⁡A=RHS\begin{aligned} &= \frac{(\cos^2 A + \sin^2 A) + 1 + 2 \sin A}{(1 + \sin A) \cos A} \\[0.6em] &= \frac{1 + 1 + 2 \sin A}{(1 + \sin A) \cos A} \\[0.6em] &= \frac{2 + 2 \sin A}{(1 + \sin A) \cos A} \\[0.6em] &= \frac{2(1 + \sin A)}{(1 + \sin A) \cos A} \\[0.6em] &= \frac{2}{\cos A} \\[0.6em] &= 2 \sec A = \text{RHS} \end{aligned}
Answer

(ii) LHS=RHS\text{LHS} = \text{RHS}

(iii) Prove tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ cosec θ\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta

Step 1 · Convert LHS to sin⁡θ\sin \theta and cos⁡θ\cos \theta

Writing in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta:

LHS=sin⁡θcos⁡θ1−cos⁡θsin⁡θ+cos⁡θsin⁡θ1−sin⁡θcos⁡θ=sin⁡θcos⁡θsin⁡θ−cos⁡θsin⁡θ+cos⁡θsin⁡θcos⁡θ−sin⁡θcos⁡θ=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)+cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ)\begin{aligned} \text{LHS} &= \dfrac{\dfrac{\sin \theta}{\cos \theta}}{1 - \dfrac{\cos \theta}{\sin \theta}} + \dfrac{\dfrac{\cos \theta}{\sin \theta}}{1 - \dfrac{\sin \theta}{\cos \theta}} \\[1em] &= \dfrac{\dfrac{\sin \theta}{\cos \theta}}{\dfrac{\sin \theta - \cos \theta}{\sin \theta}} + \dfrac{\dfrac{\cos \theta}{\sin \theta}}{\dfrac{\cos \theta - \sin \theta}{\cos \theta}} \\[1em] &= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)} \end{aligned}

Since (cos⁡θ−sin⁡θ)=−(sin⁡θ−cos⁡θ)(\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta):

LHS=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)−cos⁡2θsin⁡θ(sin⁡θ−cos⁡θ)\text{LHS} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}

Taking LCM sin⁡θcos⁡θ(sin⁡θ−cos⁡θ)\sin \theta \cos \theta (\sin \theta - \cos \theta):

=sin⁡3θ−cos⁡3θsin⁡θcos⁡θ(sin⁡θ−cos⁡θ)= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}

Using a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2) and sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1:

=(sin⁡θ−cos⁡θ)(sin⁡2θ+sin⁡θcos⁡θ+cos⁡2θ)sin⁡θcos⁡θ(sin⁡θ−cos⁡θ)=1+sin⁡θcos⁡θsin⁡θcos⁡θ=1sin⁡θcos⁡θ+sin⁡θcos⁡θsin⁡θcos⁡θ=cosec θsec⁡θ+1=1+sec⁡θ cosec θ=RHS\begin{aligned} &= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)} \\[0.6em] &= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \\[0.6em] &= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\[0.6em] &= \text{cosec } \theta \sec \theta + 1 \\[0.6em] &= 1 + \sec \theta \text{ cosec } \theta = \text{RHS} \end{aligned}
Answer

(iii) LHS=RHS\text{LHS} = \text{RHS}

(iv) Prove 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A}

Step 1 · Simplify LHS and RHS separately

Simplifying LHS\text{LHS}:

LHS=1+sec⁡Asec⁡A=1+1cos⁡A1cos⁡A=cos⁡A+1cos⁡A1cos⁡A=cos⁡A+1cos⁡A⋅cos⁡A1=1+cos⁡A\begin{aligned} \text{LHS} &= \frac{1 + \sec A}{\sec A} \\[0.6em] &= \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}} \\[1em] &= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} \\[1em] &= \frac{\cos A + 1}{\cos A} \cdot \frac{\cos A}{1} \\[0.6em] &= 1 + \cos A \end{aligned}

Simplifying RHS\text{RHS} using sin⁡2A=1−cos⁡2A\sin^2 A = 1 - \cos^2 A:

RHS=sin⁡2A1−cos⁡A=1−cos⁡2A1−cos⁡A=(1−cos⁡A)(1+cos⁡A)1−cos⁡A=1+cos⁡A\begin{aligned} \text{RHS} &= \frac{\sin^2 A}{1 - \cos A} \\[0.6em] &= \frac{1 - \cos^2 A}{1 - \cos A} \\[0.6em] &= \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} \\[0.6em] &= 1 + \cos A \end{aligned}

Since LHS=RHS=1+cos⁡A\text{LHS} = \text{RHS} = 1 + \cos A, the identity is verified.

Answer

(iv) LHS=RHS\text{LHS} = \text{RHS}

(v) Prove cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cosec A+cot⁡A\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A, using the identity cosec2A=1+cot⁡2A\text{cosec}^2 A = 1 + \cot^2 A

Step 1 · Divide by sin⁡A\sin A and apply identity

Dividing numerator and denominator by sin⁡A\sin A:

LHS=cos⁡Asin⁡A−sin⁡Asin⁡A+1sin⁡Acos⁡Asin⁡A+sin⁡Asin⁡A−1sin⁡A=cot⁡A−1+cosec Acot⁡A+1−cosec A=(cot⁡A+cosec A)−1cot⁡A−cosec A+1\begin{aligned} \text{LHS} &= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} \\[1.1em] &= \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A} \\[0.6em] &= \frac{(\cot A + \text{cosec } A) - 1}{\cot A - \text{cosec } A + 1} \end{aligned}

Substitute 1=cosec2A−cot⁡2A=(cosec A−cot⁡A)(cosec A+cot⁡A)1 = \text{cosec}^2 A - \cot^2 A = (\text{cosec } A - \cot A)(\text{cosec } A + \cot A) in the numerator:

=(cot⁡A+cosec A)−(cosec2A−cot⁡2A)cot⁡A−cosec A+1=(cot⁡A+cosec A)−(cosec A−cot⁡A)(cosec A+cot⁡A)cot⁡A−cosec A+1=(cot⁡A+cosec A)[1−(cosec A−cot⁡A)]cot⁡A−cosec A+1=(cot⁡A+cosec A)(1−cosec A+cot⁡A)cot⁡A−cosec A+1=cot⁡A+cosec A=RHS\begin{aligned} &= \frac{(\cot A + \text{cosec } A) - (\text{cosec}^2 A - \cot^2 A)}{\cot A - \text{cosec } A + 1} \\[0.6em] &= \frac{(\cot A + \text{cosec } A) - (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}{\cot A - \text{cosec } A + 1} \\[0.6em] &= \frac{(\cot A + \text{cosec } A)[1 - (\text{cosec } A - \cot A)]}{\cot A - \text{cosec } A + 1} \\[0.6em] &= \frac{(\cot A + \text{cosec } A)(1 - \text{cosec } A + \cot A)}{\cot A - \text{cosec } A + 1} \\[0.6em] &= \cot A + \text{cosec } A = \text{RHS} \end{aligned}
Answer

(v) LHS=RHS\text{LHS} = \text{RHS}

(vi) Prove 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A

Step 1 · Rationalize the denominator inside the root

Multiplying numerator and denominator by (1+sin⁡A)(1 + \sin A) under the radical:

LHS=(1+sin⁡A)(1+sin⁡A)(1−sin⁡A)(1+sin⁡A)=(1+sin⁡A)21−sin⁡2A=(1+sin⁡A)2cos⁡2A=1+sin⁡Acos⁡A=1cos⁡A+sin⁡Acos⁡A=sec⁡A+tan⁡A=RHS\begin{aligned} \text{LHS} &= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \\[0.8em] &= \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} \\[0.8em] &= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \\[0.8em] &= \frac{1 + \sin A}{\cos A} \\[0.6em] &= \frac{1}{\cos A} + \frac{\sin A}{\cos A} \\[0.6em] &= \sec A + \tan A = \text{RHS} \end{aligned}
Answer

(vi) LHS=RHS\text{LHS} = \text{RHS}

(vii) Prove sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta

Step 1 · Factor out common terms and simplify

Factoring sin⁡θ\sin \theta and cos⁡θ\cos \theta:

LHS=sin⁡θ(1−2sin⁡2θ)cos⁡θ(2cos⁡2θ−1)\begin{aligned} \text{LHS} &= \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)} \end{aligned}

Using sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta in the numerator:

=sin⁡θ[1−2(1−cos⁡2θ)]cos⁡θ(2cos⁡2θ−1)=sin⁡θ(1−2+2cos⁡2θ)cos⁡θ(2cos⁡2θ−1)=sin⁡θ(2cos⁡2θ−1)cos⁡θ(2cos⁡2θ−1)=sin⁡θcos⁡θ=tan⁡θ=RHS\begin{aligned} &= \frac{\sin \theta [1 - 2(1 - \cos^2 \theta)]}{\cos \theta (2 \cos^2 \theta - 1)} \\[0.6em] &= \frac{\sin \theta (1 - 2 + 2 \cos^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)} \\[0.6em] &= \frac{\sin \theta (2 \cos^2 \theta - 1)}{\cos \theta (2 \cos^2 \theta - 1)} \\[0.6em] &= \frac{\sin \theta}{\cos \theta} \\[0.6em] &= \tan \theta = \text{RHS} \end{aligned}
Answer

(vii) LHS=RHS\text{LHS} = \text{RHS}

(viii) Prove (sin⁡A+cosec A)2+(cos⁡A+sec⁡A)2=7+tan⁡2A+cot⁡2A(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A

Step 1 · Expand and apply trigonometric identities

Expanding using (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2:

LHS=(sin⁡2A+2sin⁡A cosec A+cosec2A)+(cos⁡2A+2cos⁡Asec⁡A+sec⁡2A)\begin{aligned} \text{LHS} &= (\sin^2 A + 2 \sin A \text{ cosec } A + \text{cosec}^2 A) + (\cos^2 A + 2 \cos A \sec A + \sec^2 A) \end{aligned}

Using sin⁡A cosec A=1\sin A \text{ cosec } A = 1 and cos⁡Asec⁡A=1\cos A \sec A = 1:

=sin⁡2A+2(1)+cosec2A+cos⁡2A+2(1)+sec⁡2A=(sin⁡2A+cos⁡2A)+4+cosec2A+sec⁡2A\begin{aligned} &= \sin^2 A + 2(1) + \text{cosec}^2 A + \cos^2 A + 2(1) + \sec^2 A \\[0.6em] &= (\sin^2 A + \cos^2 A) + 4 + \text{cosec}^2 A + \sec^2 A \end{aligned}

Using sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1, cosec2A=1+cot⁡2A\text{cosec}^2 A = 1 + \cot^2 A, and sec⁡2A=1+tan⁡2A\sec^2 A = 1 + \tan^2 A:

=1+4+(1+cot⁡2A)+(1+tan⁡2A)=7+tan⁡2A+cot⁡2A=RHS\begin{aligned} &= 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A) \\[0.6em] &= 7 + \tan^2 A + \cot^2 A = \text{RHS} \end{aligned}
Answer

(viii) LHS=RHS\text{LHS} = \text{RHS}

(ix) Prove (cosec A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A(\text{cosec } A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}

Step 1 · Simplify LHS and RHS separately

Simplifying LHS\text{LHS} using cosec A=1sin⁡A\text{cosec } A = \dfrac{1}{\sin A} and sec⁡A=1cos⁡A\sec A = \dfrac{1}{\cos A}:

LHS=(1sin⁡A−sin⁡A)(1cos⁡A−cos⁡A)=(1−sin⁡2Asin⁡A)(1−cos⁡2Acos⁡A)=(cos⁡2Asin⁡A)(sin⁡2Acos⁡A)=sin⁡Acos⁡A\begin{aligned} \text{LHS} &= \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) \\[0.6em] &= \left(\frac{1 - \sin^2 A}{\sin A}\right)\left(\frac{1 - \cos^2 A}{\cos A}\right) \\[0.6em] &= \left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right) \\[0.6em] &= \sin A \cos A \end{aligned}

Simplifying RHS\text{RHS} by expressing in sin⁡A\sin A and cos⁡A\cos A:

RHS=1tan⁡A+cot⁡A=1sin⁡Acos⁡A+cos⁡Asin⁡A=1sin⁡2A+cos⁡2Asin⁡Acos⁡A=11sin⁡Acos⁡A=sin⁡Acos⁡A\begin{aligned} \text{RHS} &= \frac{1}{\tan A + \cot A} \\[0.6em] &= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} \\[1em] &= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \\[1em] &= \frac{1}{\frac{1}{\sin A \cos A}} \\[0.6em] &= \sin A \cos A \end{aligned}

Since LHS=RHS=sin⁡Acos⁡A\text{LHS} = \text{RHS} = \sin A \cos A, the identity is proved.

Answer

(ix) LHS=RHS\text{LHS} = \text{RHS}

(x) Prove (1+tan⁡2A1+cot⁡2A)=(1−tan⁡A1−cot⁡A)2=tan⁡2A\left(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A

Step 1 · Simplify the first expression

Using 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A and 1+cot⁡2A=cosec2A1 + \cot^2 A = \text{cosec}^2 A:

1+tan⁡2A1+cot⁡2A=sec⁡2Acosec2A=1cos⁡2A1sin⁡2A=sin⁡2Acos⁡2A=tan⁡2A\begin{aligned} \frac{1 + \tan^2 A}{1 + \cot^2 A} &= \frac{\sec^2 A}{\text{cosec}^2 A} \\[0.6em] &= \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} \\[1em] &= \frac{\sin^2 A}{\cos^2 A} \\[0.6em] &= \tan^2 A \end{aligned}

Step 2 · Simplify the second expression

Converting tan⁡A\tan A and cot⁡A\cot A into sin⁡A\sin A and cos⁡A\cos A:

(1−tan⁡A1−cot⁡A)2=(1−sin⁡Acos⁡A1−cos⁡Asin⁡A)2=(cos⁡A−sin⁡Acos⁡Asin⁡A−cos⁡Asin⁡A)2=(cos⁡A−sin⁡Acos⁡A⋅sin⁡Asin⁡A−cos⁡A)2=(cos⁡A−sin⁡Acos⁡A⋅sin⁡A−(cos⁡A−sin⁡A))2=(−sin⁡Acos⁡A)2=(−tan⁡A)2=tan⁡2A\begin{aligned} \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 &= \left(\frac{1 - \frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}}\right)^2 \\[1em] &= \left(\frac{\frac{\cos A - \sin A}{\cos A}}{\frac{\sin A - \cos A}{\sin A}}\right)^2 \\[1em] &= \left(\frac{\cos A - \sin A}{\cos A} \cdot \frac{\sin A}{\sin A - \cos A}\right)^2 \\[0.8em] &= \left(\frac{\cos A - \sin A}{\cos A} \cdot \frac{\sin A}{-(\cos A - \sin A)}\right)^2 \\[0.8em] &= \left(-\frac{\sin A}{\cos A}\right)^2 \\[0.6em] &= (-\tan A)^2 \\[0.6em] &= \tan^2 A \end{aligned}

Both expressions simplify to tan⁡2A\tan^2 A, so the identity holds.

Answer

(x) LHS=RHS=tan⁡2A\text{LHS} = \text{RHS} = \tan^2 A

Common Mistakes
  • Sign Errors in Binomials: In part (iii) and part (x), failing to note (cos⁡θ−sin⁡θ)=−(sin⁡θ−cos⁡θ)(\cos \theta - \sin \theta) = -(\sin \theta - \cos \theta) causes incorrect sign cancellations.
  • Algebraic Identity Confusion: Forgetting that (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 includes the cross-term 2ab2ab, which simplifies nicely when aa and bb are reciprocals (e.g. sin⁡A cosec A=1\sin A \text{ cosec } A = 1).
  • Replacing 11 Selectively: In part (v), replacing the 11 in both the numerator and denominator creates unnecessary complexity; substituting 1=cosec2A−cot⁡2A1 = \text{cosec}^2 A - \cot^2 A only in the numerator allows direct factoring.

More questions in Exercise 8.3

Q1

Express the trigonometric ratios sin⁡A\sin A, sec⁡A\sec A and tan⁡A\tan A in terms of cot⁡A\cot A.

Q2

Write all the other trigonometric ratios of ∠A\angle A in terms of sec⁡A\sec A.

Q3

Choose the correct option. Justify your choice.

(i) 9sec⁡2A−9tan⁡2A=9 \sec^2 A - 9 \tan^2 A =

(A) 11 (B) 99 (C) 88 (D) 00

(ii) (1+tan⁡θ+sec⁡θ)(1+cot⁡θ−csc⁡θ)=(1 + \tan \theta + \sec \theta) (1 + \cot \theta - \csc \theta) =

(A) 00 (B) 11 (C) 22 (D) −1-1

(iii) (sec⁡A+tan⁡A)(1−sin⁡A)=(\sec A + \tan A) (1 - \sin A) =

(A) sec⁡A\sec A (B) sin⁡A\sin A (C) csc⁡A\csc A (D) cos⁡A\cos A

(iv) 1+tan⁡2A1+cot⁡2A=\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =

(A) sec⁡2A\sec^2 A (B) −1-1 (C) cot⁡2A\cot^2 A (D) tan⁡2A\tan^2 A

Q4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ(\text{cosec } \theta - \cot \theta)^2 = \dfrac{1 - \cos \theta}{1 + \cos \theta}

(ii) cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡A\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2 \sec A

(iii) tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θ cosec θ\dfrac{\tan \theta}{1 - \cot \theta} + \dfrac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta [Hint : Write the expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta]

(iv) 1+sec⁡Asec⁡A=sin⁡2A1−cos⁡A\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A} [Hint : Simplify LHS and RHS separately]

(v) cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cosec A+cot⁡A\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \text{cosec } A + \cot A, using the identity cosec2A=1+cot⁡2A\text{cosec}^2 A = 1 + \cot^2 A.

(vi) 1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A

(vii) sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta

(viii) (sin⁡A+cosec A)2+(cos⁡A+sec⁡A)2=7+tan⁡2A+cot⁡2A(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A

(ix) (cosec A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A(\text{cosec } A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A} [Hint : Simplify LHS and RHS separately]

(x) (1+tan⁡2A1+cot⁡2A)=(1−tan⁡A1−cot⁡A)2=tan⁡2A\left(\dfrac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A

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