Question

Prove that,

sec²A + cot² (90° - A) = 2 cosec² (90° - A) - 1

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#Cubingwitsk

Answer 1

$\bold{Answer :}$

L.H.S. = sec²A + cot² (90° - A)

= (1 + tan²A) + [cosec² (90° - A) - 1],

using the identities
sec²θ - tan²θ = 1, cosec²θ - cot²θ = 1

= 1 + tan²A + cosec² (90° - A) - 1

= tan²A + cosec² (90° - A)

= (tanA)² + cosec² (90° - A)

= {cot (90° - A)}² + cosec² (90° - A),

since cot (90 - θ) = tanθ

= {cosec² (90° - A) - 1} + cosec² (90° - A),

using the identity
cosec²θ - cot²θ = 1

= 2 cosec² (90° - A) - 1

= R.H.S.

Hence, proved.

#$\bold{MarkAsBrainliest}$

Answer 2

♥️Hi there ♥️

sec²A + cot² (90° - A) = 2 cosec² (90° - A) - 1

From L.H.S

${sec}^{2} a + {cot}^{2} (90 - a) \\ \\ = {sec}^{2} a + {tan}^{2} a$

(Cot(90°-a)= tan A

${sec}^{2} a + ( {sec}^{2} a - 1)$

(sec^2 A =tan^2A=1)

$= 2 {sec}^{2} a - 1$

From R.H.S

$2 {cosec}^{2} (90 - a) - 1$

(Cosec(90°- A) = secA

$= 2 {sec}^{2} a - 1$

L.H.S = R.H.S

Hence proved !

Thanks !!

【In my ans a=A and in few places 90 = 90°】