Advertisements
Advertisements
Question
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Advertisements
Solution
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
`sin(50^circ + θ) = cos[90^circ - (50^circ + θ)] = cos(40^circ - θ)`
`sin(50^circ + θ) - cos(40^circ - θ)`
= `cos(40^circ - θ) - cos(40^circ - θ)`
= 0
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
If x = r sin θ cos Φ, y = r sin θ sin Φ and z = r cos θ, prove that x2 + y2 + z2 = r2.
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
