Trigonometric Functions — MATHS STD 11 Science — Question

3. $\frac{\pi}{6}$

Solution:

We have:

$\tan\text{x}+\sec\text{x}=\sqrt{3}$ $[0,<\text{x}<\pi]$

$\Rightarrow\sec\text{x}+\tan\text{x}=\sqrt{3}$

$\Rightarrow\frac{1}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}=\sqrt{3}$

$\Rightarrow1+\sin\text{x}=\sqrt{3}\cos\text{x}$

$\Rightarrow(1+\sin\text{x})^2=(\sqrt{3}\cos\text{x})^2$

$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3\cos^2\text{x}$

$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3(1-\sin^2\text{x})$

$\Rightarrow4\sin^2\text{x}+2\sin\text{x}=2$

$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$

$\Rightarrow\sin\text{x}=-1,\frac{1}{2}$

since $0<\text{x}<\pi,\ \sin\text{x}$ cannot be negative.

$\therefore\sin\text{x}=\frac{1}{2}$

$\therefore\text{x}=\frac{\pi}{6}$