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प्रश्न
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
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उत्तर
L.H.S. = `tan^2A - tan^2B`
= `sin^2A/cos^2A - sin^2B/cos^2B`
= `(sin^2A * cos^2B - sin^2B * cos^2A)/(cos^2A * cos^2B`
= `(sin^2A(1 - sin^2B)-sin^2B(1 - sin^2A))/(cos^2A * cos^2B)`
= `(sin^2A - sin^2A * sin^2B - sin^2B + sin^2A * sin^2B)/(cos^2A * cos^2B`
= `(sin^2A - cancel(sin^2A * sin^2B) - sin^2B + cancel(sin^2A * sin^2B))/(cos^2A * cos^2B`
= `(sin^2A - sin^2B)/(cos^2A * cos^2B)` = R.H.S.
संबंधित प्रश्न
Prove that: `(1 – sinθ + cosθ)^2 = 2(1 + cosθ)(1 – sinθ)`
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
cosec4θ − cosec2θ = cot4θ + cot2θ
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
