Advertisements
Advertisements
प्रश्न
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
Advertisements
उत्तर
L.H.S. = `(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2`
= `(sinA/cosA + 1/cosA)^2 + (sinA/cosA - 1/cosA)^2`
= `((sinA + 1)/cosA)^2 + ((sinA - 1)/cosA)^2`
= `(sinA + 1)^2/(cos^2A) + (sinA - 1)^2/(cos^2A)`
= `((sinA + 1)^2 + (sinA - 1)^2)/(cos^2A)`
= `(sin^2A + 1 + 2sinA + sin^2A + 1 - 2sinA)/cos^2A`
= `(2sin^2A + 2)/(cos^2A)`
= `(2(1 + sin^2A))/(1 - sin^2A)`
= `2((1 + sin^2A)/(1 - sin^2A))` = R.H.S.
संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove the following identity :
`(1 + cotA)^2 + (1 - cotA)^2 = 2cosec^2A`
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Choose the correct alternative:
cos 45° = ?
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
