Advertisements
Advertisements
प्रश्न
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
Advertisements
उत्तर
L.H.S. = `sinAcosA - (sinAsinAcosA)/(cosecA) - (cosAcosAsinA)/secA`
= sin A cos A – sin2 A cos A sin A – cos2 A sin A cos A
= sin A cos A – sin3 A cos A – cos3 A sin A
= sin A cos A [1 – sin2 A – cos2 A]
= sin A cos A [1 – (sin2 A + cos2 A)]
= sin A cos A (1 – 1)
= sin A cos A × 0
= 0 = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
Find the value of sin ` 48° sec 42° + cos 48° cosec 42°`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
