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प्रश्न
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
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उत्तर
L.H.S. = (sec A − tan A)2 (1 + sin A)
`(1/cos "A" - sin "A"/cos "A")^2 (1 + sin "A")`
= `((1 - sin "A")/cos "A")^2 (1 + sin "A")`
= `((1 - sin "A")(1 - sin "A")(1 + sin "A"))/cos^2"A"`
= `((1 - sin "A")(1 - sin^2 "A"))/cos^2"A"`
= `((1 - sin "A")cos^2"A")/cos^2"A"`
= (1 − sin A) R.H.S.
संबंधित प्रश्न
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
If `secθ = 25/7 ` then find tanθ.
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove the following identities: sec2 θ + cosec2 θ = sec2 θ cosec2 θ.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
