Advertisements
Advertisements
प्रश्न
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Advertisements
उत्तर
LHS = secA(1 + sinA)(secA - tanA)
= `1/cosA(1 + sinA)(1/cosA - sinA/cosA)`
= `((1 + sinA))/cosA((1-sinA)/cosA) = (1-sin^2A)/cos^2A`
= `(cos^2A/cos^2A) = 1` = RHS
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}`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
If ` cot A= 4/3 and (A+ B) = 90° ` ,what is the value of tan B?
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
