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प्रश्न
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
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उत्तर
LHS = cosec2 (90° - θ) + cot2 (90° - θ)
= sec2 θ + tan2θ
= 1 + tan2θ + tan2θ
= 1 + 2 tan2θ
= RHS
Hence proved.
संबंधित प्रश्न
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove that `sec"A"/(tan "A" + cot "A")` = sin A
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
