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प्रश्न
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
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उत्तर
LHS = `(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)`
= `(1-1)/((cosA + cosB)(sinA - sinB))`
= `0/((cosA + cosB)(sinA - sinB))`
= 0
`(sinA + sinB)/(cosA + cosB) + (cosA - cosB)/(sinA - sinB) = 0`
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संबंधित प्रश्न
Prove the following trigonometric identities
`cos theta/(1 - sin theta) = (1 + sin theta)/cos theta`
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
If `sin theta = x , " write the value of cot "theta .`
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos 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.
