Advertisements
Advertisements
Question
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Advertisements
Solution
R.H.S = `(1/(sin A) + 1)/(1/(sin A) - 1)`
= `((1 + sin A)/sin A)/((1 - sin A)/sin A)`
= `((1 + sin A))/cancelsin A xx cancelsin A/((1 - sin A))`
= `(1 + sin A)/(1 - sin A)`
∴ R.H.S = L.H.S
RELATED QUESTIONS
Prove the following trigonometric identities.
if `T_n = sin^n theta + cos^n theta`, prove that `(T_3 - T_5)/T_1 = (T_5 - T_7)/T_3`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
(1 + sin A)(1 – sin A) is equal to ______.
