# Inverse Trigonometric Function Questions

## The value of $$tan^{-1}(1)$$ + $$cos^{-1}({-1\over 2})$$ + $$sin^{-1}({-1\over 2})$$ is equal to

Solution : We have, $$tan^{-1}(1)$$ + $$cos^{-1}({-1\over 2})$$ + $$sin^{-1}({-1\over 2})$$ = $$\pi\over 4$$ + $$2\pi\over 3$$ – $$\pi\over 6$$ = $$3\pi\over 4$$ Similar Questions Solve the equation : 2$$tan^{-1}({2x+1})$$ = $$cos^{-1}x$$ Prove that : $$sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$ Evaluate $$sin^{-1}(sin10)$$ Prove that : $$cos^{-1}{12\over 13}$$ + $$sin^{-1}{3\over 5}$$ …

Solution : Here, 2$$tan^{-1}({2x+1})$$ = $$cos^{-1}x$$ cos(2$$tan^{-1}({2x+1})$$) = x  { We Know cos2x = $${1-tan^2x\over {1+tan^2x}}$$} $$\therefore$$  $${{1-{(2x+1)}^2}\over {1-{(2x+1)}^2}}$$ = x   $$\implies$$   (1 – 2x – 1)(1 + 2x + 1) = x($$4x^2 + 4x + 2$$) $$\implies$$  -2x.2(x + 1) = 2x($$2x^2 + 2x + 1$$)  $$\implies$$  2x($$2x^2 + 2x + 1 + 2x … ## Prove that : \(sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$

Solution : We have, $$sin^{-1}{12\over 13}$$ + $$cot^{-1}{4\over 3}$$ + $$tan^{-1}{63\over 16}$$ = $$tan^{-1}{12\over 5}$$ + $$tan^{-1}{3\over 4}$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$ + $$tan^{-1}({{{12\over 5} + {3\over 4}}\over {1 – {12\over 5} \times {3\over 4}}})$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$ + $$tan^{-1}{63\over (-16)}$$ + $$tan^{-1}{63\over 16}$$ = $$\pi$$ – $$tan^{-1}{63\over 16}$$ + $$tan^{-1}{63\over 16}$$ …

## Evaluate $$sin^{-1}(sin10)$$

Solution : We know that $$sin^{-1}(sinx)$$ = x, if $$-\pi\over 2$$ $$\le$$ x $$\le$$ $$\pi\over 2$$ Here, x = 10 radians which does not lie between -$$\pi\over 2$$ and $$\pi\over 2$$ But, $$3\pi$$ – x i.e. $$3\pi$$ – 10 lie between -$$\pi\over 2$$ and $$\pi\over 2$$ Also, sin($$3\pi$$ – 10) = sin 10 $$\therefore$$  $$sin^{-1}(sin10)$$ …

## Prove that : $$cos^{-1}{12\over 13}$$ + $$sin^{-1}{3\over 5}$$ = $$sin^{-1}{56\over 65}$$

Solution : We have, L.H.S. = $$cos^{-1}{12\over 13}$$ + $$sin^{-1}{3\over 5}$$ = $$tan^{-1}{5\over 12}$$ + $$tan^{-1}{3\over 4}$$ $$\because$$ [ $$cos^{-1}{12\over 13}$$ = $$tan^{-1}{5\over 12}$$ & $$sin^{-1}{3\over 5}$$ = $$tan^{-1}{3\over 4}$$ ] L.H.S. = $$tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})$$ = $$tan^{-1}{56\over 33}$$ R.H.S. = $$sin^{-1}{56\over 65}$$ = $$tan^{-1}{56\over 33}$$ L.H.S = …

## Find the value of $$sin^{-1}({-\sqrt{3}\over 2})$$ + $$cos^{-1}(cos({7\pi\over 6}))$$.

Solution : $$sin^{-1}({-\sqrt{3}\over 2})$$ = – $$sin^{-1}({\sqrt{3}\over 2})$$ = $$-\pi\over 3$$ $$cos^{-1}(cos({7\pi\over 6}))$$ = $$cos^{-1}(cos({2\pi – {5\pi\over 6}}))$$ = $$cos^{-1}(cos({5\pi\over 6}))$$ = $$5\pi\over 6$$ Hence $$sin^{-1}({-\sqrt{3}\over 2})$$ + $$cos^{-1}(cos({7\pi\over 6}))$$ = $$-\pi\over 3$$ + $$5\pi\over 6$$ = $$\pi\over 2$$ Similar Questions Solve the equation : 2$$tan^{-1}({2x+1})$$ = $$cos^{-1}x$$ Prove that : $$sin^{-1}{12\over 13}$$ + \(cot^{-1}{4\over …