NDA 2014
Exam NDA
Year 2014
Subject Mathematics
Chapter Inverse Trigonometric Functions NDA 2014 · Mathematics · Inverse Trigonometric Functions
What is ^ -1 (3)/(5)+ ^ -1 (4)/(5) equal to?
What is
sin − 1 3 5 + sin − 1 4 5 \sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5} sin − 1 5 3 + sin − 1 5 4 equal to?
Options A π 2 \frac{\pi}{2} 2 π B π 3 \frac{\pi}{3} 3 π C π 4 \frac{\pi}{4} 4 π D π 6 \frac{\pi}{6} 6 π Correct answer A. π 2 \frac{\pi}{2} 2 π
Step-by-step solution
Let
sin − 1 3 5 = x ⇒ sin x = 3 5 ∴ cos x = 4 5 \sin ^{-1} \frac{3}{5}=x\Rightarrow \quad \sin x=\frac{3}{5} \qquad \therefore \cos x=\frac{4}{5} sin − 1 5 3 = x ⇒ sin x = 5 3 ∴ cos x = 5 4 and
sin − 1 4 5 = y ⇒ sin y = 4 5 ∴ cos y = 3 5 \quad \sin ^{-1} \frac{4}{5}=y\Rightarrow \quad \sin y=\frac{4}{5} \qquad \therefore \cos y=\frac{3}{5} sin − 1 5 4 = y ⇒ sin y = 5 4 ∴ cos y = 5 3 sin ( x + y ) = sin x ⋅ cos y + cos x ⋅ sin y = 3 5 × 3 5 + 4 5 × 4 5 = 9 25 + 16 25 = 25 25 = 1 ∴ sin ( x + y ) = sin π 2 ⇒ x + y = π 2 ⇒ sin − 1 3 5 + sin − 1 4 5 = π 2 . \begin{aligned} \sin (x+y) & =\sin x \cdot \cos y+\cos x \cdot \sin y \\ & =\frac{3}{5} \times \frac{3}{5}+\frac{4}{5} \times \frac{4}{5}=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1 \\ \therefore \quad \sin (x+y) & =\sin \frac{\pi}{2} \\ \Rightarrow \quad x+y & =\frac{\pi}{2} \\ \Rightarrow & \sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}.\end{aligned} sin ( x + y ) ∴ sin ( x + y ) ⇒ x + y ⇒ = sin x ⋅ cos y + cos x ⋅ sin y = 5 3 × 5 3 + 5 4 × 5 4 = 25 9 + 25 16 = 25 25 = 1 = sin 2 π = 2 π sin − 1 5 3 + sin − 1 5 4 = 2 π . $$
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