Question

Good Evening Friends

Please solve the question !!!

CLASS -X
MATHS
CH - TRIGONOMETRY

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Answer 1

Here, I am writing θ as A for better understanding!.

$=>Given:\frac{1-cosA}{1+cosA}$

$=>\frac{1-cosA}{1+cosA}*\frac{1-cosA}{1-cosA}$

$=>\frac{(1-cosA)^2}{1^2-cos^2A}$

$=>\frac{1+cos^2A-2cosA}{1-cos^2A}$

$=>\frac{1+cos^2A-2cosA}{sin^2A}$

$=>\frac{1}{sin^2A}+\frac{cos^2A}{sin^2A}-\frac{2cosA}{sin^2A}$

$=>cosec^2A+cot^2A-\frac{2}{sinA}* \frac{cosA}{sinA}$

$=>cosec^2A+cot^2A-2cosecAcotA$

$=>\boxed{(cosecA-cotA)^2}$

Hope it helps!

Answer 2

Given Equation is (1 - cosθ)/(1+cosθ)

On rationalizing, we get

⇒ (1 - cosθ/1 + cosθ) * (1 - cosθ/1 - cosθ)

⇒ (1 - cosθ)^2/sin^2θ

⇒ (1 + cos^2θ-2cosθ)/sin^2θ

⇒ (1/sin^2θ) + (cos^2θ/sin^2θ) - (2cosθ/sinA)

⇒ cosec^2θ + cot^2θ - 2cosecθcotθ

⇒ (cosecθ - cotθ)^2.

Hope this helps u!