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

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