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प्रश्न
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
उत्तर
`1/(cosA + sinA) + 1/(cosA - sinA)`
= `(cosA + sinA + cosA - sinA)/((cosA + sinA)(cosA - sinA)`
= `(2cosA)/(cos^2A - sin^2A)`
= `(2cosA)/(cos^2A - (1 - cos^2A))`
= `(2cosA)/(cos^2A - 1 + cos^2A)`
= `(2cosA)/(2cos^2A - 1)`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`
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Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
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If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
