SATHEE: Inverse Circular Functions 3 Question 1

Inverse Circular Functions 3 Question 1

1. The value of $\sin^{- 1} \frac{12}{13} - \sin^{- 1} \frac{3}{5}$ is equal to

(a) $π - \sin^{- 1} \frac{63}{65}$

(b) $\frac{π}{2} - \sin^{- 1} \frac{56}{65}$

(d) $π - \cos^{- 1} \frac{33}{65}$

(2019 Main, 12 April I)

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Answer:

Correct Answer: 1. (b)

Solution:

$\begin{aligned} Key Idea Use formulae \\ (i) \sin^{- 1} x - \sin^{- 1} y \\ = \sin^{- 1} (x \sqrt{1 - y^{2}} - y \sqrt{1 - x^{2}}) if x^{2} + y^{2} \leq 1 \\ or if x y > 0 and x^{2} + y^{2} > 1 \forall x, y \in [- 1, 1] \\ (ii) \sin^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}} and \\ (iii) \sin^{- 1} θ + \cos^{- 1} θ = \frac{π}{2} \end{aligned}$

We have,

$\begin{aligned} \sin^{- 1} \frac{12}{13} & - \sin^{- 1} \frac{3}{5} \\ = \sin^{- 1} \frac{12}{13} \sqrt{1 - {\frac{3}{5}}^{2}} - \frac{3}{5} \sqrt{1 - {\frac{12}{13}}^{2}} \end{aligned}$

$[∵ \sin^{- 1} x - \sin^{- 1} y = \sin^{- 1} (x \sqrt{1 - y^{2}} - y \sqrt{1 - x^{2}})$ ,

if $x^{2} + y^{2} \leq 1$ or if $x y > 0$ and $x^{2} + y^{2} > 1 \forall x, y \in [- 1, 1]]$

$= \sin^{- 1} \frac{12}{13} \times \frac{4}{5} - \frac{3}{5} \times \frac{5}{13}$

$= \sin^{- 1} \frac{48 - 15}{65}$

$= \sin^{- 1} \frac{33}{65}$

$= \cos^{- 1} \sqrt{1 - \frac{33}{65}}$

$= \cos^{- 1} \sqrt{\frac{3136}{4225}}$ $∵ \sin^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}}$

$= \cos^{- 1} \frac{56}{65} = \frac{π}{2} - \sin^{- 1} \frac{56}{65}$ $∵ \sin^{- 1} θ + \cot^{- 1} θ = \frac{π}{2}$

Inverse Circular Functions 3 Question 1

1. The value of $\sin^{- 1} \frac{12}{13} - \sin^{- 1} \frac{3}{5}$ is equal to

Answer:

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Inverse Circular Functions 3 Question 1

1. The value of sin−1⁡1213−sin−1⁡35 is equal to

Answer:

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1. The value of $\sin^{- 1} \frac{12}{13} - \sin^{- 1} \frac{3}{5}$ is equal to