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प्रश्न
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
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उत्तर
LHS = ( 1 + tan A)2 + (1 - tan A)2
= 1 + 2 tan A + tan2A + 1 - 2 tan A + tan2A
= 2( 1 + tan2A)
= 2 sec2A
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
Prove the following identities:
cot2 A – cos2 A = cos2 A . cot2 A
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
Show that none of the following is an identity:
`sin^2 theta + sin theta = 2`
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
If 2sin2θ – cos2θ = 2, then find the value of θ.
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`
