Question

$\sqrt{cot { }^{2} - cos {}^{2} } = cot .cos$
evaluate the following trigonometric equation. ​

Answer 1

√(cot^2-cos^2) = cot cos.

squaring both the sides

(cot^2-cos^2) = cot^2 Cos^2

(cot-cos) (cot+cos) = cot^2Cos^2

((cos/sin)-cos) ((cos/sin)+cos) = Cot^2 cos ^ 2

((cos-cos. Sin)/ Sin)(cos+cos. Sin)/sin=cot^2 Cos^2

(cos ^2-(Cos. Sin)^2)/Sin^2=Cos^4/Sinn ^2

(cos^2(1-Sin^2))/Sin^2=tan^2.cos^2

tan^2 Cos^2 = tan^2 Cos^2

LHS=RHS

Thus,

proved.