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प्रश्न
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
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उत्तर
L.H.S. = `(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB)`
= `((sinA - sinB)(sinA + sinB) + (cosA - cosB)(cosA + cosB))/((cosA + cosB)(sinA + sinB)`
= `(sin^2A - sin^2B + cos^2A - cos^2B)/((cosA + cosB)(sinA + sinB))`
= `((sin^2A + cos^2A) - (sin^2B + cos^2B))/((cosA + cosB)(sinA + sinB))` ...[∵ cos2 A + sin2 A = 1]
= `(1 - 1)/((cosA + cosB)(sinA + sinB))`
= 0 = R.H.S.
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संबंधित प्रश्न
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Prove the following trigonometric identities.
`sin^2 A + 1/(1 + tan^2 A) = 1`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
Write the value of `(1+ tan^2 theta ) ( 1+ sin theta ) ( 1- sin theta)`
Prove the following identity :
`sec^2A + cosec^2A = sec^2Acosec^2A`
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
Prove that (sec θ + tan θ) (1 – sin θ) = cos θ
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`
