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Question
Prove the following identities:
`(1 + (secA - tanA)^2)/(cosecA(secA - tanA)) = 2tanA`
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Solution
`(1 + (secA - tanA)^2)/(cosecA(secA - tanA))`
= `((sec^2A - tan^2A) + (secA - tanA)^2)/(cosecA(secA - tanA))`
= `((secA - tanA)(secA + tanA) + (secA + tanA)^2)/(cosecA(secA - tanA))`
= `((secA + tanA) + (secA - tanA))/(cosecA)`
= `(2secA)/(cosecA)`
= `2(1/cosA)/(1/sinA)`
= 2 tanA
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RELATED QUESTIONS
Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
If `secθ = 25/7 ` then find tanθ.
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
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
