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Correct CHOICE is (b) \([\frac{-1}{3}, \frac{1}{3}]\)

The EXPLANATION: The domain of y=sin^-1⁡x is -1≤x≤1.

∴the domain of y=sin^-1⁡3x is-1≤3x≤1

⇒ \([\frac{-1}{3} ≤ x ≤ \frac{1}{3}]\)

Hence, \([\frac{-1}{3}, \frac{1}{3}]\).

The CORRECT OPTION is (a) sin^-1⁡(\(\frac{63}{65}\))

Best explanation: From ∆ABC, we get

cos^-1⁡\((\frac{3}{5})\)=sin^-1⁡\((\frac{4}{5})\)

∴sin^-1⁡(\(\frac{5}{13}\))+cos^-1⁡(\(\frac{3}{5}\))=sin^-1⁡(\(\frac{5}{13}\))+sin^-1⁡(\(\frac{4}{5}\))

=sin^-1⁡\((\frac{5}{13}\SQRT{1-(\frac{4}{5})^2}+\frac{4}{5}\sqrt{1-(\frac{5}{13})^2})\)

=\(sin^{-1}(\frac{5}{13}×\frac{3}{5}+\frac{4}{5}×\frac{12}{13})=sin^{-1}(\frac{15+48}{65})=sin^{-1}(\frac{63}{65})\).

The CORRECT ANSWER is (a) sin^-1⁡x

Explanation: [-1, 1] is the DOMAIN for sin^-1⁡x.

The domain for cot^-1⁡x is (-∞,∞).

The domain for tan^-1⁡x is (-∞,∞).

The domain for sec^-1⁡x is (-∞,-1]∪[1,∞).

The correct OPTION is (C) 2 sec^-1⁡\((\sqrt{1+x^2})\)

The EXPLANATION: Let 2 tan^-1⁡x=y

⇒tan^-1⁡x=\(\frac{y}{2}\)

From ∆ABC, we get

⇒tan^-1⁡x=sec^-1⁡\(\sqrt{1+x^2}=\frac{y}{2}\)

⇒y=2 sec^-1⁡(\(\sqrt{1+x^2}\))

Correct OPTION is (d) –\(\frac{π}{3}\)

To elaborate: \(SIN^{-1}⁡(sin⁡X)\)=x, x∈\([-\frac{π}{2},\frac{π}{2}]\)

∴\(sin^{-1} (sin⁡ \frac{4π}{3})=sin^{-1}⁡(sin⁡(π+\frac{π}{3}))=sin^{-1}⁡(sin⁡(\frac{-π}{3}))= -\frac{π}{3}\).

Correct OPTION is (a) -SIN^-1(x)

The BEST EXPLANATION: Let, θ = sin^-1(-x)

So, -π/2 ≤ θ ≤ π/2

=> -x = sinθ

=> x = -sinθ

=> x = sin(-θ)

Also, -π/2 ≤ -θ ≤ π/2

=> -θ = sin^-1(x)

=> θ = -sin^-1(x)

So, sin^-1(-x) = -sin^-1(x)

The CORRECT answer is (c) π

For EXPLANATION: \(TAN^{-1}⁡\FRAC{1}{\sqrt{3}}=\frac{π}{6},sin^{-1}⁡1=\frac{π}{2}, cos^{-1}\frac{⁡1}{2}=\frac{π}{3}\)

\(tan^1⁡\frac{1}{\sqrt{3}}-sin^{-1}⁡1+ cos^{-1}⁡\frac{1}{2}=\frac{π}{6}+\frac{π}{2}+\frac{π}{3}=\frac{π+3π+2π}{6}=\frac{6π}{6}=π\)

The correct answer is (B) sin^-1⁡x-sin^-1⁡y=sin^-1⁡{\(x\SQRT{1+y^2}+y\sqrt{1+x^2}\)}

EASY EXPLANATION: The formula sin^-1⁡x-sin^-1⁡y=sin^-1⁡{\(x\sqrt{1+y^2}+y\sqrt{1+x^2}\)} is incorrect. The correct formula is sin^-1⁡x-sin^-1⁡y=sin^-1⁡{\(x\sqrt{1+y^2}-y \sqrt{1-x^2}\)}.

Correct choice is (a) 0 ≤ y ≤ π

To elaborate: Given that, \(cos^{-1}⁡x=y\)

The RANGE of PRINCIPLE values for the inverse trigonometric function \(cos^{-1}\) is [0,π].

Hence, 0≤y≤π.

Right option is (B) \(\frac{2π}{3}\)

Easy explanation: USING the formula sin^-1⁡X+sin^-1⁡y=sin^-1⁡\({x \sqrt{1-y^2}+y \sqrt{1-x^2}}\), we get

sin^-1⁡(\(\frac{3}{5}\))+sin^-1⁡(\(\frac{4}{5}\))=sin^-1\(\Big\{ \frac{3}{5} \sqrt{1-(\frac{4}{5})^2}+\frac{4}{5} \sqrt{1-(\frac{3}{5})^2} \Big\}\)

=sin^-1⁡(\(\frac{3}{5}\)×\(\frac{3}{5}\)+\(\frac{4}{5}\)×\(\frac{4}{5}\))=sin^-1⁡\((\frac{25}{25})=\frac{π}{2}\)

∴ sin^-1⁡(\(\frac{3}{5}\))+sin^-1⁡(\(\frac{4}{5}\))+cos^-1⁡\((\frac{\sqrt{3}}{2})=\frac{π}{2}+\frac{π}{6}=\frac{3π+π}{6}=\frac{2π}{3}\).

Right answer is (B) \(\FRAC{2π}{3}\)

The explanation: \(TAN^{-1}\sqrt{3}=\frac{π}{3}, sec^{-1}⁡2=\frac{π}{3}, COS^{-1}⁡1=0\)

∴\(tan^{-1}\sqrt{3}+sec^{-1}⁡2 -cos^{-1}⁡1=\frac{π}{3}+\frac{π}{3}-0\)

=\(\frac{2π}{3}\).

The CORRECT OPTION is (a) \(\FRAC{π}{2}\)

The EXPLANATION: sin^-1⁡x+cos^-1⁡x=\(\frac{π}{2}\); x∈[-1,1]