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Question: If \(\sin \theta\), \(\cos \theta\) and \(\tan \theta\) are in G.P. then find the value of \(\cot^{6...

If sin⁡θ\sin \theta, cos⁡θ\cos \theta and tan⁡θ\tan \theta are in G.P. then find the value of cot⁡6θ−cot⁡2θ\cot^{6} \theta − \cot^{2} \theta.

A. 1

B. 12

C. 2

D. 3

Explanation

Solution

Hint: Use the basic definition of GP and the necessary trigonometric identities to arrive at the answer.

Complete step-by-step answer:

Now, we know that if 3 terms a, b, c are in GP then we can write,

ba=cb=r\dfrac{b}{a} = \dfrac{c}{b} = r, where r is the common ratio.

The above can also be written as

⇒b2=ac→(1)\Rightarrow b^2 = ac \to (1)

It is given that sin⁡θ\sin \theta, cos⁡θ\cos \theta, tan⁡θ\tan \theta are in GP, so from eqn (1), we can write,

cos⁡2θ=sin⁡θ×tan⁡θ→(2)\cos^{2} \theta = \sin \theta \times \tan \theta \to (2)

Now tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{\sin \theta}{\cos \theta}, substituting in equation (2), we get,

⇒cos⁡2θ=sin⁡θ×sin⁡θcos⁡θ\Rightarrow \cos^{2} \theta = \sin \theta \times {\dfrac{\sin \theta}{\cos \theta}}

⇒cos⁡2θ=sin⁡2θcos⁡θ\Rightarrow \cos^{2} \theta = \dfrac{\sin^{2} \theta}{\cos \theta}

⇒sin⁡2θcos⁡3θ=1→(3)\Rightarrow \dfrac{\sin^{2} \theta}{\cos^{3} \theta} =1 \to (3)

It can be also rearranged as,

⇒cos⁡2θsin⁡2θ=1cos⁡θ=sec⁡θ\Rightarrow \dfrac{\cos^{2} \theta}{\sin^{2} \theta} = \dfrac{1}{\cos \theta} = \sec \theta

⇒cot⁡2θ=sec⁡θ→(4)(∵cos⁡θsin⁡θ=cot⁡θ)\Rightarrow \cot^{2} \theta = \sec \theta \to (4) (\because \dfrac{\cos \theta}{\sin \theta} = \cot \theta)

Now, we have to find the value of cot⁡6θ−cot⁡2θ\cot^{6} \theta - \cot^{2} \theta

It can be written as, (cot⁡2θ)3−cot⁡2θ(\cot^{2} \theta)^3 - \cot^{2} \theta

Substituting from eqn (4), we get,

⇒sec⁡3θ−sec⁡θ\Rightarrow \sec^{3} \theta - \sec \theta

Taking sec⁡θ\sec \theta common

⇒sec⁡θ(sec⁡2θ−1)\Rightarrow \sec \theta (\sec^{2} \theta - 1)

We know (from trigonometric identities) that sec⁡2θ−1=tan⁡2θ\sec^{2} \theta - 1 = \tan^{2} \theta, we get

⇒sec⁡θ×tan⁡2θ\Rightarrow \sec \theta \times \tan^{2} \theta

⇒1cos⁡θ×sin⁡2θcos⁡2θ\Rightarrow \dfrac{1}{\cos \theta} \times \dfrac{\sin^{2} \theta}{\cos^{2} \theta}

⇒sin⁡2θcos⁡3θ\Rightarrow \dfrac{\sin^{2} \theta}{\cos^{3} \theta}

=1= 1 (from eqn. (3))

Hence the value of cot⁡6θ−cot⁡2θ\cot^{6} \theta - \cot^{2} \theta is 1.

∴\therefore Option A. is correct.

Note: Such problems, where more than one concept is involved can be solved easily by knowing the basics of those concepts. Using the necessary properties will solve these problems. Mistakes can be avoided while rearranging and substituting.