Exercise 4C

### Solve for :

Solution: Given: $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ We know that $\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}$ So, $\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x$ Substituting in the...

### Solve for :

Solution: To find: value of $x$ Given: $\sin ^{-1} \frac{\mathrm{g}}{\mathrm{x}}+\sin ^{-1} \frac{15}{\mathrm{x}}=\frac{\pi}{2}$ We know $\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}$...

### Solve for x:

Solution: To find: value of $x$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where xy $<1$ Given: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$...

### Prove that:

Solution: To Prove: $\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi$ Formula Used: $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\pi+\tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)$...

### Prove that:

Solution: To Prove: $\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} \frac{1}{3}$ Formula Used: $\tan ^{-1} \mathrm{x}-\tan ^{-1} \mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{x}-\mathrm{y}}{1+\mathrm{xy}}\right)$...