Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

$\cos A,\sin A,\cot A$ are in $G.P$

$\therefore {\sin }^{2}A=\cos A\times \cot A=\cos A\times \frac{\cos A}{\sin A}\Rightarrow {\sin }^{2}A=\frac{{\cos }^{2}A}{\sin A}\Rightarrow {\sin }^{3}A={\cos }^{2}A-\left(i\right){\tan }^{6}A-{\tan }^{2}A=\frac{{\sin }^{6}A}{{\cos }^{6}A}-\frac{{\sin }^{2}A}{{\cos }^{2}A}=\frac{{\left({\sin }^{3}A\right)}^2}{{\cos }^{6}A}-\frac{(1-{\cos }^{2}A)}{{\cos }^{2}A}=\frac{{\cos }^{4}A}{{\cos }^{6}A}-\frac{1}{{\cos }^{2}A}+1=\frac{1}{{\cos }^{2}A}-\frac{1}{{\cos }^{2}A}+1=1$