Question

If the terms $\sin a,\cos a,\tan a$ are in G.P. then ${{\cot }^{6}}a-{{\cot }^{2}}a$ is equal to: -
(a) 1
(b) -1
(c) 0
(d) 2

Answer 1

Apply the formula to find the geometric mean of three terms x, y, z given as ${{y}^{2}}=xz$. Form relation between trigonometric functions by using the conversion, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\dfrac{1}{\cos \theta }=\sec \theta$. Substitute the value of $\cot a$ in the given expression and simplify it to get the answer. Use the trigonometric identity: - ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta$.

Complete step-by-step solution
Here, we have been provided with the information that $\sin a,\cos a,\tan a$ are three terms in G.P.
Now, we know that if three terms like x, y, z are in G.P. then its geometric mean is given by the relation: - ${{y}^{2}}=xz$. Therefore, applying this formula, we get,
$\Rightarrow {{\cos }^{2}}a=\sin a\times \tan a$
Using the conversion: - $\tan a=\dfrac{\sin a}{\cos a}$ we get,

& \Rightarrow {{\cos }^{2}}a=\sin a\times \dfrac{\sin a}{\cos a} \\\ & \Rightarrow {{\cos }^{2}}a=\dfrac{{{\sin }^{2}}a}{\cos a} \\\ \end{aligned}$$ Dividing both sides by $${{\sin }^{2}}a$$, we get, $$\Rightarrow \dfrac{{{\cos }^{2}}a}{{{\sin }^{2}}a}=\dfrac{1}{\cos a}$$ Using the conversion, $$\dfrac{\cos \theta }{\sin \theta }=\cot \theta $$ and $$\dfrac{1}{\cos \theta }=\sec \theta $$, we get, $$\Rightarrow {{\cot }^{2}}a=\sec a$$ - (1) Now, we have to find the value of expression: - $${{\cot }^{6}}a-{{\cot }^{2}}a$$. Let us assume its value as ‘E’. $$\Rightarrow E={{\cot }^{6}}a-{{\cot }^{2}}a={{\left( {{\cot }^{2}}a \right)}^{3}}-{{\cot }^{2}}a$$ Substituting the value of $${{\cot }^{2}}a$$ from equation (1), we get, $$\begin{aligned} & \Rightarrow E={{\left( \sec a \right)}^{3}}-\sec a \\\ & \Rightarrow E=\sec a\left[ {{\sec }^{2}}a-1 \right] \\\ \end{aligned}$$ Using the trigonometric identity: - $${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta $$, we get, $$\Rightarrow E=\sec a\times {{\tan }^{2}}a$$ - (2) From equation (1) we have, $$\Rightarrow {{\cot }^{2}}a=\sec a$$ Taking reciprocal both sides, we get, $$\begin{aligned} & \Rightarrow \dfrac{1}{{{\cot }^{2}}a}=\dfrac{1}{\sec a} \\\ & \Rightarrow {{\tan }^{2}}a=\cos a \\\ \end{aligned}$$ So, substituting the value of $${{\tan }^{2}}a$$ from the above relation in equation (2), we get, $$\Rightarrow E=\sec a\times \cos a$$ We know that, $$\sec a\times \cos a=1$$, since they are reciprocal of each other. Therefore, we have, $$\begin{aligned} & \Rightarrow E=1 \\\ & \Rightarrow {{\cot }^{6}}a-{{\cot }^{2}}a=1 \\\ \end{aligned}$$ **Hence, option (a) is the correct answer.** **Note:** One may note that the power of $$\cot a$$ in the assumed expression ‘E’ is 6 which is a large number, therefore our main objective was to reduce the power by finding the relation between trigonometric functions. You must remember the formula of the geometric mean of three numbers and some trigonometric identities to solve the above question. Remember the geometric mean is different from the arithmetic mean, so do not apply the formula for the arithmetic mean of three numbers otherwise we will get the wrong answer.