If x= 1 5 , y= 1 10 , where 0 < x < (π)/(2), 0 < y < (π)/(2) then what is (x+y) equal to?

Correct answer: C. (π)/(4) aligned & x= 1 5 , y= 1 10 , 0 < x < (π)/(2), 0 < y < (π)/(2) & x= 1- ^2 x & y= 1- ^2 y & x= 1-(1)/(5) & y= 1-(1)/(10) aligned aligned & x= (4)/(5) = 2 5 & y= (9)/(10) = 3 10 & (x+y)= x y+ x y & = 1 5 · 3 10 + 2 5 · 1 10 & = 5 5 · 10 = 5 · 5 5 · 10 = (5)/(10) = (1)/(2) = 1 2 & x+y= ^ -1 ( 1 2 )=(π)/(4) aligned

NDA 2018

Exam: NDA
Year: 2018
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

If x= 1 5 , y= 1 10 , where 0 < x < (π)/(2), 0 < y < (π)/(2) then...

\sin x=\frac{1}{\sqrt{5}}, \sin y=\frac{1}{\sqrt{10}}

, where

0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2}

then what is

(x+y)

equal to?

Options

A $\pi$
B $\frac{\pi}{2}$
C $\frac{\pi}{4}$
D0

Correct answer

C. $\frac{\pi}{4}$

Step-by-step solution

\begin{aligned} & \sin x=\frac{1}{\sqrt{5}}, \sin y=\frac{1}{\sqrt{10}}, 0 < x < \frac{\pi}{2}, 0 < y < \frac{\pi}{2} \\ & \because \cos x=\sqrt{1-\sin ^2 x} \quad & \cos y=\sqrt{1-\sin ^2 y} \\ & \Rightarrow \cos x=\sqrt{1-\frac{1}{5}} \quad & \quad \cos y=\sqrt{1-\frac{1}{10}}\end{aligned}

\begin{aligned} & \Rightarrow \cos x=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}} \quad & \cos y=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}} \\ & \sin (x+y)=\sin x \cos y+\cos x \sin y \\ & =\frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{10}} \\ & =\frac{5}{\sqrt{5} \cdot \sqrt{10}}=\frac{\sqrt{5} \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{10}}=\sqrt{\frac{5}{10}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \\ & \therefore \quad x+y=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\end{aligned}

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