Question

tan^2thita-sin^2 thita =tan^2thita.sin^2thita​

Answer 1

Answer:

you mean,we have to prove (tan²θ - sin²θ) = tan²θ × sin²θ , right ?

LHS = tan²θ - sin²θ

= sin²θ/cos²θ - sin²θ

= sin²θ [ 1/cos²θ - 1]

[ we know, secx = 1/cosx so, 1/cos²θ = sec²θ]

= sin²θ [ sec²θ - 1]

we also know, sec²x - tan²x = 1

so, sec²θ - tan²θ = 1

or, sec²θ - 1 = tan²θ

then, sin²θ[sec²θ - 1] = sin²θ × tan²θ = RHS

hence, (tan²θ - sin²θ) = tan²θ × sin²θ

hope this helps you

thank you!

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