Advertisements
Advertisements
Question
Prove that `( 1 + sin θ)/(1 - sin θ) = 1 + 2 tan θ/cos θ + 2 tan^2 θ` .
Advertisements
Solution
RHS = `1 + 2 tan θ/cos θ + 2 tan^2 θ`
= `1 + 2 sin θ/cos^2θ + 2 sin^2 θ/cos^2 θ`
= `(cos^2 θ + 2sin θ + 2 sin^2 θ)/(cos^2θ)`
= `(1 - sin^2θ + 2 sin θ + 2 sin^2θ )/(1 - sin^2θ)`
= `(1 + sin^2θ + 2 sin θ)/(1 - sin^2θ)`
= `(1 + sin θ)^2/( 1 + sin θ)(1 - sin θ)`
= `(1 + sin θ)/(1 - sin θ)`
= LHS
Hence proved.
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
(sec A + tan A) (1 − sin A) = ______.
If A = 30°, verify that `sin 2A = (2 tan A)/(1 + tan^2 A)`.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
