Advertisements
Advertisements
प्रश्न
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
उत्तर
`cos theta/(1 + sin theta)` = sec θ – tan θ
R.H.S. = sec θ – tan θ
= `1/cos theta - sin theta/cos theta`
= `(1 - sin theta)/costheta`
= `(1 - sin theta)/cos theta xx (1 + sin theta)/(1 + sin theta)`
= `(1 - sin^2 theta)/(cos theta(1 + sin theta)`
= `cos^2 theta/(cos theta(1 + sin theta))`
= `costheta/(1 + sintheta)`
L.H.S. = R.H.S.
∴ `cos theta/(1 + sin theta)` = sec θ – tan θ
Aliter:
L.H.S. = `cos theta/(1 - sin theta)` ...[conjugate (1 – sin θ)]
= `(cos theta(1 + sin theta))/((1 - sin theta)(1 + sin theta))`
= `(cos theta(1 + sin theta))/((1 - sin^2 theta))`
= `(cos theta (1 + sin theta))/(cos^2 theta)`
= `(1 + sin theta)/costheta`
L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 - sin theta)/(1 + sin theta) = (sec theta - tan theta)^2`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Without using trigonometric table , evaluate :
`(sin49^circ/sin41^circ)^2 + (cos41^circ/sin49^circ)^2`
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Choose the correct alternative:
cot θ . tan θ = ?
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.