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Question
Prove that:
Solution
L.H.S = \[\sec^4 A\left( 1 - \sin^4 A \right) - 2 \tan^2 A\]
\[ = \sec^4 A - \sec^4 A \sin^4 A - 2 \tan^2 A\]
= `(sec^2"A")^2 - 1/(cos^4"A"). sin^4"A" - 2tan^2"A" ...[secθ = 1/cosθ]`
= `(1 + tan^2"A")^2 - (sin^4"A")/(cos^4"A") - 2 tan^2"A" ...[1 + tan^2θ = sec^2θ]`
= `1^2 + 2 xx 1 xx tan^2"A" + (tan^2"A")^2 - tan^4"A" - 2tan^2"A" ...[(a + b)^2 = a^2 + 2ab + b^2 tanθ = (sinθ)/(cosθ)]`
= `1 + cancel(2tan^2"A") + cancel(tan^4"A") - cancel(tan^4"A") - cancel(2tan^2"A")`
= 1
= R.H.S
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Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
