Advertisements
Advertisements
Question
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Advertisements
Solution
LHS = `(sec θ - 1)/(sec θ + 1)`
= `(1/cos θ - 1)/(1/cos θ + 1)`
= `(1 - cos θ)/(1 + cos θ)`
= `(1 - cos θ xx ( 1 + cos θ))/(1 + cos θ xx (1 + cos θ))`
= `(1 - cos^2 θ)/(1 + cos θ)^2`
= `(sin^2 θ)/(1 + cos θ)^2`
= `((sin θ)/(1 + cos θ ))^2`
= RHS
RELATED QUESTIONS
(secA + tanA) (1 − sinA) = ______.
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
What is the value of \[\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}\]
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`.
`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.
