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प्रश्न
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
विकल्प
cot2 A
cot2 B
−tan2 A
−cot2 A
उत्तर
We have:
`A+B=90°`
`⇒ B=90°-A`
We have to find the value of the following expression
`(tan A tan B+tan A cot B)/(sin A sec B)-sin ^2 B/cos^2 A`
so
`(tan A tan B+tan A cot B )/(sin A sec B)- sin^2 B/cos^2 A`
`(tan A tan (90°-A)+ tan A cot (90°-A))/(sin A sec (90°-A))-sin^2(90°-A)/cos^2 A`
=` (tan A cot A + tan A tan A)/(sin A cosec A)-cos^2A/cos^2 A`
= `1+tan^2 A-1`
=` tan ^2 A`
=` tan ^2(90°-B)`
=` cot^2 B`
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