Jitender Gupta and Divya Malik Solutions for Exercise 1: CHAPTER EXERCISE

If $2 \sin \theta -1=0$, then prove that $\sec \theta +\tan \theta =\sqrt{3}$.

Prove that: $\frac{\sin \theta }{1-cot\theta }+\frac{\cos {\displaystyle \theta }}{1-\tan \theta }=\cos \theta +\sin \theta$

Prove that $\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{{\displaystyle {\sin }^2\mathrm{A}}}{1-\cos \mathrm{A}}$.

Without using trigonometric table, evaluate:

${\cos }^2 {26}^0 + \cos {64}^0 \sin {26}^0 + \frac{\tan {36}^0}{\cot {54}^0}$

Without using trigonometric table, evaluate.

$2{\left(\frac{\tan 35°}{\cot 55°}\right)}^2+{\left(\frac{\cot 55°}{\tan 35°}\right)}^2-3\left(\frac{\sec 40°}{\mathrm{cosec} 50°}\right)$

Without using trigonometric tables, evaluate:

$3 \cos {80}^0 \mathrm{cosec} {10}^0 + 2 \sin {59}^0 \sec {31}^0$

Without using trigonometric table, evaluate:

${\sin }^2 {34}^0 + {\sin }^2 {56}^0 + 2 t\mathrm{an} {18}^0 \tan {72}^0 - co{t}^2 {30}^0$

Evaluate: 

$\frac{{\sec }^2({90}^0-\theta )-{\cot }^2 \theta }{2\left({\sin }^2 {25}^0+{\sin }^2 {65}^0\right)} + \frac{2\left({\cos }^2 {60}^0 {\tan }^2 {28}^0 {\tan }^2 {62}^0\right)}{\sin {\displaystyle }{\displaystyle {30}^0}{\displaystyle }{\displaystyle \cos }{\displaystyle }{\displaystyle {60}^0}}$