Advertisements
Advertisements
Question
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Advertisements
Solution
LHS = `(sin θ/cos θ + sin θ)/(sin θ/cos θ - sin θ)`
= `(sin θ (1/cos θ + 1))/(sin θ (1/cos θ - 1))`
= `(sec θ + 1)/(sec θ - 1)`
= RHS
Hence proved.
RELATED QUESTIONS
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
`(1-cos^2theta) sec^2 theta = tan^2 theta`
`1/((1+tan^2 theta)) + 1/((1+ tan^2 theta))`
`sqrt((1+sin theta)/(1-sin theta)) = (sec theta + tan theta)`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
