Question

If x + y = ${90}^{\circ }$, then what is the value of $(1+\frac{\tan x}{\tan y}){\sin }^{2}y$?

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

The correct option is C 1

$x+y=90°\Rightarrow x=90°-y$

$\Rightarrow (1+\frac{\tan x}{\tan y}){\sin }^{2}y=(1+\frac{\tan (90°-y)}{\tan y})-{\sin }^{2}y$

$=(1+\frac{\cot y}{\tan y}){\sin }^{2}y$

$=(1+{\cot }^{2}y){\sin }^{2}y$

$=cose{c}^{2}y{\sin }^{2}y$

$=\frac{1}{{\sin }^{2}y}{\sin }^{2}y$

$=1$

Hence, the answer is $1.$