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प्रश्न
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
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उत्तर
L.H.S. = `(1 + sinA)/cosA + cosA/(1 + sinA)`
= `((1 + sinA)^2 + cos^2A)/(cosA(1 + sinA))`
= `(1 + sin^2A + 2sinA + cos^2A)/(cosA(1 + sinA))`
= `(1 + 2sinA + 1)/(cosA(1 + sinA))`
= `(2(1 + sinA))/(cosA(1 + sinA))`
= 2 sec A = R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Prove the following identity :
`(tanθ + sinθ)/(tanθ - sinθ) = (secθ + 1)/(secθ - 1)`
Find the value of ( sin2 33° + sin2 57°).
tan θ cosec2 θ – tan θ is equal to
Choose the correct alternative:
1 + cot2θ = ?
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
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`
