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प्रश्न
Prove that:
(sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2
Prove the following:
(sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2
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उत्तर
(sin A + sec A)2 + (cos A + cosec A)2 = 1 + (sec A + cosec A)2
L.H.S = [(sin A + sec A)2 + (cos A + cosec A)2]
= [sin2 A + sec2 A + 2 sin A sec A + cos2 A + cosec2 A + 2 cos A cosec A]
= (sin2A + cos2A) + (sec2A + cosec2A) + 2 (sin A sec A + cos A cosec A)
= `(sin^2 A + cos^2 A) + (sec^2 A + cosec^2 A) + 2[sin A xx 1/cos A + cos A xx 1/sin A]`
= `1 + (1/cos^2 A + 1/sin^2 A) + 2(sin A/cos A + cos A/sin A)`
= `1 + (sin^2 A+ cos^2 A)/(sin^2 A cos^2 A) + 2 ((sin^2 A + cos^2 A )/(sin A cos A)) ...[∵ sin^2 A + cos^2 A = 1]`
= `1 + 1/(sin^2 A cos^2 A) + 2 xx 1/((sin A cos A))`
= 1 + (sec2A cosec2A) + 2 sec A cosec A
= 1 + 2 sec A cosec A + (sec A cosec A)2 ...(∵ Rearrange as a2 + 2ab + b2.)
= (1 + sec A cosec A)2
L.H.S = R.H.S
∴ (sin A + sec A)2 + (cos A + cosec A)2 = (1 + sec A cosec A)2
