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Question: If \(\tan\theta - \cot\theta = a\) and \(\sin\theta + \cos\theta = b,\) then \((b^{2} - 1)^{2}(a^{2}...

If tan⁡θ−cot⁡θ=a\tan\theta - \cot\theta = a and sin⁡θ+cos⁡θ=b,\sin\theta + \cos\theta = b, then (b2−1)2(a2+4)(b^{2} - 1)^{2}(a^{2} + 4) is equal to

A

2

B

– 4

C

± 4

D

4

Answer

4

Explanation

Solution

Given that tan⁡θ−cot⁡θ=a\tan\theta - \cot\theta = a …..(i)

and sin⁡θ+cos⁡θ=b\sin\theta + \cos\theta = b …..(ii)

Now (b2−1)2(a2+4)(b^{2} - 1)^{2}(a^{2} + 4)

={(sin⁡θ+cos⁡θ)2−1}2{(tan⁡θ−cot⁡θ)2+4}= \left\{ (\sin\theta + \cos\theta)^{2} - 1 \right\}^{2}\left\{ (\tan\theta - \cot\theta)^{2} + 4 \right\}

=[1+sin⁡2θ−1]2[tan⁡2θ+cot⁡2θ−2+4]= \lbrack 1 + \sin 2\theta - 1\rbrack^{2}\lbrack\tan^{2}\theta + \cot^{2}\theta - 2 + 4\rbrack

=sin⁡22θ(cosec2θ+sec⁡2θ)= \sin^{2}2\theta(\text{cose}\text{c}^{2}\theta + \sec^{2}\theta)

=4sin⁡2θcos⁡2θ[1sin⁡2θ+1cos⁡2θ]=4= 4\sin^{2}\theta\cos^{2}\theta\left\lbrack \frac{1}{\sin^{2}\theta} + \frac{1}{\cos^{2}\theta} \right\rbrack = 4.

Trick : Obviously the value of expression (b2−1)2(a2+4)(b^{2} - 1)^{2}(a^{2} + 4) is independent of θ\theta, therefore put any suitable value of θ\theta. Let θ=45∘\theta = 45{^\circ}, we get a=0, 6mub=2a = 0,\mspace{6mu} b = \sqrt{2}so that [(2)2−1]2\lbrack(\sqrt{2})^{2} - 1\rbrack^{2} (02+4)=4.(0^{2} + 4) = 4.