Advertisements
Advertisements
प्रश्न
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Advertisements
उत्तर
LHS = sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ )
= sec θ. sec θ - tan θ. tan θ
= sec2θ - tan2θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
`(1 + cot^2A)/(1 + tan^2A)` = ?
Let α, β be such that π < α – β < 3π. If sin α + sin β = `-21/65` and cos α + cos β = `-27/65`, then the value of `cos (α - β)/2` is ______.
