Advertisements
Advertisements
प्रश्न
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
Advertisements
उत्तर
LHS = sin( 90° - θ ) sin θ cot θ
= cos θ . sin θ . `cos θ/sin θ`
= cos2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
Prove that: sin4 θ + cos4θ = 1 - 2sin2θ cos2 θ.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
If `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
