Advertisements
Advertisements
प्रश्न
Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`
Advertisements
उत्तर
LHS = `( sin θ tan θ)/(1 - cos θ)`
= `(sin θ. (sin θ)/(cos θ))/(1 - cos θ)`
= `sin^2 θ/(cos θ( 1 - cos θ))`
= `((1 - cos θ)(1 + cos θ))/(cos θ(1 - cos θ))`
= `(1 + cos θ)/(cos θ) = 1/(cos θ) + cos θ/cos θ`
= sec θ + 1
= RHS
Hence proved.
संबंधित प्रश्न
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
Prove that:
Sin4θ - cos4θ = 1 - 2cos2θ
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 identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
If sin θ + sin2 θ = 1 show that: cos2 θ + cos4 θ = 1
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
