Advertisements
Advertisements
प्रश्न
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
Advertisements
उत्तर
L.H.S. = `(1 - sinA)/(1 + sinA)`
= `(1 - sinA)/(1 + sinA) xx (1 - sinA)/(1 - sinA)`
= `(1 - sinA)^2/(1 - sin^2A`
= `(1 - sinA)^2/cos^2A`
= `((1 - sinA)/cosA)^2`
= `(1/cosA - sinA/cosA)^2`
= `(secA - tanA)^2`
= R.H.S.
APPEARS IN
संबंधित प्रश्न
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}`
Four alternative answers for the following question are given. Choose the correct alternative and write its alphabet:
sin θ × cosec θ = ______
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
If cos A + cos2A = 1, then sin2A + sin4A = ?
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
