Question

proove that : tan^2alpha-tan^2beta = sec^2alpha-sec^2beta = sin^2alpha-sin^2beta /cos^2alpha-cos^2beta
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Answer 1

Answer:

Step-by-step explanation:

First solve each term individually.

tan²α-tan²β = sec²α-sec²β

⇒tan²α-sec²α= tan²β-sec²β ------------ A

We know : 1 + tan²∅ = sec²∅

∴  sec²∅ - tan²∅ = 1

MULTIPLYING BOTH SIDES BY "-1"

tan²∅ -sec²∅ = 1  ---------------- (i)

Now using 1 in eqn A we get

-1 = -1

Now solving last term

sin²α-sin²β / cos²α - cos²β ---------B

We know that:

sin²Ф + cos²Ф = 1

∴ sin²Ф = 1- cos²Ф ---------- (ii)

Now using (ii) in B

1-cos²α-1+cos²β/ cos²α-cos²β

⇒-cos²α+cos²β/ cos²α-cos²β

⇒-1

∴ All 3 terms are equal

Hence proved