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प्रश्न
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
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उत्तर
LHS = sec2 (90° - θ) + tan2 (90° - θ)
= cosec2θ + cot2θ
= 1 + cot2θ + cot2θ
= 1 + 2cot2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
