Question

If $x=sin\left(2ta{n}^{-1}2\right),y=sin\left(\frac{1}{2}ta{n}^{-1}\frac{4}{3}\right)$ then

• A. x = 1 – y
• B. $x^2=1–y$
• C. $x^2=1+y$
• D. $y^2=1-x$

Answer 1

The correct option is D $y^2=1-x$

Let $ta{n}^{-1}2=\alpha \Rightarrow x=sin2\alpha =\frac{4}{5}y=sin\left(\frac{\beta }{2}\right)=\sqrt{\frac{1-cos\beta }{2}}=\sqrt{\frac{1-\frac{3}{5}}{2}}=\frac{1}{\sqrt{5}}$