Question

If $\cot A+\tan A=1$ then the value of ${\cot }^{2}A+{\tan }^{2}A$ is equal to

• A. $1$
• B. $2$
• C. $-2$
• D. $-1$

Answer 1

The correct option is D $-1$

$\Rightarrow$ We have, $cotA+tanA=1$

$\Rightarrow$ $\frac{cosA}{sinA}+\frac{sinA}{cosA}=1$

$\therefore$ $\frac{si{n}^{2}A+co{s}^{2}A}{sinA.cosA}=1$

$\therefore$ $sinA.cosA=1$ $[\because si{n}^{2}x+co{s}^{2}x=1]$

$\therefore$ $si{n}^{2}A.co{s}^{2}A=1$ --- ( 1 ) [ squaring both the sides ]

$\Rightarrow$ Now, $co{t}^{2}A+ta{n}^{2}A$

$\Rightarrow$ $\frac{co{s}^{2}A}{si{n}^{2}A}+\frac{si{n}^{2}A}{co{s}^{2}A}$

$\Rightarrow$ $\frac{si{n}^{4}A+co{s}^{4}A}{si{n}^{2}A.co{s}^{2}A}$

$\Rightarrow$ $\frac{(si{n}^{2}A+co{s}^{2}A{)}^2-2si{n}^{2}A.co{s}^{2}A}{1}$ [ From ( 1 ) ]

$\Rightarrow$ $\left(1\right)-2\times 1$

$\Rightarrow$ $1-2$

$\therefore$ $-1$