Question

Evaluate without using trigonometric tables:

${\sin }^2{28}^o+{\sin }^2{62}^o+{\tan }^2{38}^o-{\cot }^2{52}^o+\frac{1}{4}se{c}^2{30}^o$

Answer 1

Rearrange the equation as follows

$({\sin }^{2}28+{\sin }^{2}62)+{\tan }^{2}38-{\cot }^{2}52+\frac{1}{4}{\sec }^{2}30$

$=({\sin }^{2}28+{\cos }^2(90-28\left)\right)+{\tan }^{2}38-{\cot }^{2}52+\frac{1}{4}{\sec }^{2}30$

$=({\sin }^{2}28+{\cos }^{2}28)+{\tan }^{2}38-{\cot }^{2}52+\frac{1}{4}{\sec }^{2}30$

$=1+{\tan }^{2}38-{\cot }^{2}52+\frac{1}{4}{\sec }^{2}30$

NOTE:$1+ta{n}^2\theta =se{c}^2\theta$

$={\sec }^{2}38-{\cot }^{2}52+\frac{1}{4}{\sec }^{2}30$

$={\csc }^{2}52-{\cot }^{2}52+\frac{1}{4}{\sec }^{2}30$

$=1+\frac{1}{4}{\sec }^{2}30sec30=\frac{2}{\sqrt{3}}$

$=1+\frac{1}{4}\left(\frac{4}{3}\right)$

$=\frac{4}{3}$