Advertisements
Advertisements
प्रश्न
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Advertisements
उत्तर
LHS = cosec2 (90° - θ) + cot2 (90° - θ)
= sec2 θ + tan2θ
= 1 + tan2θ + tan2θ
= 1 + 2 tan2θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Show that none of the following is an identity:
`sin^2 theta + sin theta =2`
If m = ` ( cos theta - sin theta ) and n = ( cos theta + sin theta ) "then show that" sqrt(m/n) + sqrt(n/m) = 2/sqrt(1-tan^2 theta)`.
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Evaluate:
`(tan 65°)/(cot 25°)`
1 + cot2θ = ?
(sec θ + tan θ) . (sec θ – tan θ) = ?
If tan θ = 3, then `(4 sin theta - cos theta)/(4 sin theta + cos theta)` is equal to ______.
