verify: (1 + tan^2x)/(tan^2x) = csc^2x

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(1 + tan^2x)/(tan^2x)=

(1/tan^2x)+(tan^2x/tan^2x)=

sec^2x+1=

csc^2x=csc^2x

We know the popular trigonometric identity,

Sin^2x+cos^2x = 1, Dividing by cos^2x both sides, we get:

(sinx/cosx)^2+1 = 1/cos^2. Or

tan^2x+1 = sec^2 x. Dividing both sides by tan^2x,

(tan^2x+1)/tan^2x = sec^2x/Tan^2x = (1/cos^2x)(cos^2x/sin^2x) = 1/sin^2x = sec^2x.

L:H:S ≡ (1 + tan²x)/tan²x

**⇒ 1 + tan²θ = sec²θ**

= sec²x/tan²x

=(1/cos²x) ÷ (sin²x/cos²x)

= 1/sin²x

= cosec²x

= R:H:S

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