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प्रश्न
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
उत्तर
L.H.S. = `1/(1+cosA)+1/(1-cosA)`
= `(1 - cosA + 1 + cosA)/((1 + cosA)(1 - cosA))`
= `2/(1 - cos^2A)` ...(∵ 1 – cos2 A = sin2 A)
= `2/(sin^2A)`
= 2 cosec2 A = R.H.S.
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संबंधित प्रश्न
Prove the following trigonometric identities.
tan2θ cos2θ = 1 − cos2θ
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Choose the correct alternative:
Which is not correct formula?
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
(sec2 θ – 1) (cosec2 θ – 1) is equal to ______.
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`
