Advertisements
Advertisements
प्रश्न
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
उत्तर
(sin 67° + cos 75°)
= (sin (90°−23°) + cos (90°−15°)) .....(∵ sin(90°−θ) = cosθ and cos(90°−θ) = sinθ)
= (cos 23°+ sin 15°)
संबंधित प्रश्न
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
(secA + tanA) (1 − sinA) = ______.
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
Prove the following trigonometric identities.
`(1 - sin theta)/(1 + sin theta) = (sec theta - tan theta)^2`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
cosec4 A – cosec2 A = cot4 A + cot2 A
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that sin4A – cos4A = 1 – 2cos2A
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.