Question

If $\sin \alpha ,\cos \alpha ,\tan \alpha$ are in GP, GP means ${\cos }^2\alpha =\sin \alpha \tan \alpha$, then ${\cot }^6\alpha -{\cot }^2\alpha =$ ?

• A. $1$
• B. $0$
• C. $4$
• D. $2$

Answer 1

Answer: (A)

$1$

⁡${\cos }^2\alpha =\sin \alpha \tan \alpha$

${\cos }^3\alpha ={\sin }^2\alpha$

${\cos }^6\alpha ={\sin }^4\alpha$

${\cot }^6\alpha -{\cot }^2\alpha$

$\Rightarrow \frac{{\cos }^6\alpha }{{\sin }^6\alpha }-{\cot }^2\alpha$

$\Rightarrow \frac{1}{{\sin }^2\alpha }-\frac{{\cos }^2\alpha }{{\sin }^2\alpha }=\frac{1-{\cos }^2\alpha }{{\sin }^2\alpha }=\frac{{\sin }^2\alpha }{{\sin }^2\alpha }=1$