Advertisements
Advertisements
प्रश्न
Prove the following identities:
`1/(1 - sinA) + 1/(1 + sinA) = 2sec^2A`
Advertisements
उत्तर
L.H.S. = `1/(1 - sinA) + 1/(1 + sinA)`
= `(1 + sinA + 1 - sinA)/((1 - sinA)(1 + sinA))`
= `2/(1 - sin^2A)`
= `2/cos^2A`
= 2 sec2 A = R.H.S.
APPEARS IN
संबंधित प्रश्न
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Prove the following identities:
`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
If \[\cos A = \frac{7}{25}\] find the value of tan A + cot A.
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°.
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
(1 – cos2 A) is equal to ______.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
