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
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
If `sec theta + tan theta = x," find the value of " sec theta`
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that `(1 + tan^2 A)/(1 + cot^2 A)` = sec2 A – 1
