Question

If tan²x = 1-a² , prove that secx + tan³x cosecx = (2-a²)^3/2.​

Answer 1

Answer:

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Step-by-step explanation:

tan²θ=1-a²

LHS

=secθ+tan³θcosecθ

=√(1+tan²θ)+tan²θ×tanθ×√(1+cot²θ)

[∵, sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]

=√1+(1-a²)+(1-a²)×√(1-a²)×√{1+(1/tan²θ)}

=√(2-a²)+(1-a²)×√(1-a²)×√{1+1/(1-a²)}

=√(2-a²)+(1-a²)×√(1-a²)×√{(1-a²+1)/(1-a²)}

=√(2-a²)+(1-a²)×√(2-a²)

=√(2-a²)×(1+1-a²)

=√(2-a²)×(2-a²)

=(2-a²)¹/²⁺¹

=(2-a²)³/²

=RHS (Proved)

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Answer 2

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We have ,

secx + tan ³ X cosecx

taking sec x common.

sec x ( 1 + tan³x cosec X/ sec x)

We know that sec²x = 1+tan²x

√1 + tan²x ( 1 + tan³x cotx )

√1 + tan²x ( 1 + tan³x 1/tan x)

√1+tan ²x ( 1 + tan² x)

(1+tan²x)½( 1+tan²x)

(1+tan²x)¹+½

(1+tan²x )^³/²

tan ²x = 1-a²,so

(1+1-a²) ³/² =. ( 2 -a²)³/²

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