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Question
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
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Solution
L.H.S. = `(sec A - 1)/(sec A + 1)`
= `(1/(cosA) - 1/1)/(1/(cosA) + 1/1`
= `((1 - cos A)/cos A)/((1 + cos A)/cos A)`
= `(1 - cos A)/cos A xx cos A/(1 + cos A)`
= `(1 - cosA)/(1 + cosA)`
= R.H.S.
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(tan theta)/(1-cot theta) + (cot theta)/(1-tan theta) = 1+secthetacosectheta`
[Hint: Write the expression in terms of sinθ and cosθ]
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
