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प्रश्न
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
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उत्तर
L.H.S. = cos A (1 + cot A) + sin A (1 + tan A)
= `cosA(1 + cosA/sinA) + sinA(1 + sinA/cosA)`
= `(cosA(sinA + cosA))/sinA + (sinA(cosA + sinA))/cosA`
= `(sinA + cosA)[cosA/sinA + sinA/cosA]`
= `(sinA + cosA)[(cos^2A + sin^2A)/(sinAcosA)]`
= `(sinA + cosA) xx 1/(sinAcosA)`
= `(sinA + cosA)/(sinAcosA)` ...[∵ cos2θ + sin2θ = 1]
= `sinA/(sinAcosA) + cosA/(sinAcosA)`
= `1/cosA + 1/sinA`
= sec A + cosec A = R.H.S.
संबंधित प्रश्न
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
If sec θ = x + `1/(4"x"), x ≠ 0,` find (sec θ + tan θ)
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
sin(45° + θ) – cos(45° – θ) is equal to ______.
