If x = s i n ( 2 t a n − 1 2 ) , y = s i n ( 1 2 t a n − 1 4 3 ) then Options: A . &nbsp x = 1 – y B . &nbsp x 2 = 1 – y C . &nbsp x 2 = 1 + y D . &nbsp y 2 = 1 − x

Answer: Option D : D Let t a n − 1 2 = α ⇒ x = s i n 2 α = 4 5 y = s i n ( β 2 ) = √ 1 − c o s β 2 = √ 1 − 3 5 2 = 1 √ 5

If x=sin(2tan−12),y=sin(12tan−143) then

Question -

If $x = s i n (2 t a n^{- 1} 2), y = s i n (\frac{1}{2} t a n^{- 1} \frac{4}{3})$ then

Options:

A . x = 1 – y

B .

x^{2} = 1 - y

C .

x^{2} = 1 + y

D .

y^{2} = 1 - x

Answer: Option D
:
D
Let

t a n^{- 1} 2 = α \Rightarrow x = s i n 2 α = \frac{4}{5} y = s i n (\frac{β}{2}) = \sqrt{\frac{1 - c o s β}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \frac{1}{\sqrt{5}}

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