Advertisements
Advertisements
प्रश्न
Prove that:
`(sin^3 theta + cos^3 theta)/(sin theta + cos theta) = 1 - sin theta cos theta`
उत्तर
LHS = `(sin^3 theta + cos^3 theta)/(sin theta + cos theta)`
`= ((sin theta + cos theta)(sin^2 theta - sin theta cos theta + cos^2 theta))/(sin theta + cos theta)` ...[∵ a3 + b3 = (a + b)(a2 - ab + b2)]
= sin2 θ - sin θ cos θ + cos2 θ
= 1 - sin θ cos θ ...[sin2 θ + cos2 θ = 1]
= RHS
Hence Proved.
APPEARS IN
संबंधित प्रश्न
In the below given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20√3 m long. Find the sun’s altitude.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`((1+tan^2A)/(1+cot^2A))=((1-tanA)/(1-cotA))^2=tan^2A`
Without using trigonometric tables, prove that:
cos275° + cos215° = 1
Prove that:
`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`
Prove that:
sin θ cos (90° - θ ) + sin (90° - θ) cos θ = 1
Prove that:
cot12° cot38° cot52° cot60° cot78° = \[\frac{1}{\sqrt{3}}\]
Given that `tan (θ_1 + θ_2) = (tan θ_1 + tan θ_2)/(1 - tan θ_1 tan θ_2)` Find (θ1 + θ2) when tan θ1 = `1/2 and tan θ_2 = 1/3`.
Solve the following equation: `(cos^2θ - 3 cosθ + 2)/sin^2θ` = 1.
Using trigonometric table evaluate the following:
sin 64°42' + cos 42°20'
