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Question
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
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Solution
LHS = `(sin^2θ)/(cosθ) + cosθ = secθ`
= `(sin^2θ + cos^2θ)/(cosθ)`
= `1/(cosθ)` ...(sin2θ + cos2θ = 1)
= secθ ...`(1/cosθ = secθ)`
R.H.S
LHS = RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
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`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
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2 sin2 A + cos4 A = 1 + sin4 A
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`(tan A + tanB )/(cot A + cot B) = tan A tan B`
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Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that `"cot A"/(1 - cot"A") + "tan A"/(1 - tan "A")` = – 1
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
