Advertisements
Advertisements
प्रश्न
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
उत्तर
LHS = `cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°)`
= `cos^2 26° + cos (90° - 26°) sin 26° + (tan 36°)/(cot (90° - 54°)`
= `cos^2 26° + sin 26°. sin 26° + (tan 36°)/(tan 36°)`
= cos2 26° + sin2 26 + 1 ....( cos2 θ + sin2 θ = 1)
= 1 + 1 = 2
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
`1/(1 + cosA) + 1/(1 - cosA) = 2cosec^2A`
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove the following identities:
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0