Advertisements
Advertisements
प्रश्न
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Advertisements
उत्तर
L.H.S = `(cosec A - sin A)(secA - cosA)sec^2A`
`= (1/sinA - sinA)(1/cosA - cosA)(1/cos^2A)`
`= ((1 - sin^2A)/sin A)((1- cos^2A)/cos A)(1/(cos^2A))`
`= cos^2A/sinA . sin^2A/cos A . 1/cos^2A`
`= sinA/cosA`
= tan A
= R.H.S
APPEARS IN
संबंधित प्रश्न
Prove that ` \frac{\sin \theta -\cos \theta +1}{\sin\theta +\cos \theta -1}=\frac{1}{\sec \theta -\tan \theta }` using the identity sec2 θ = 1 + tan2 θ.
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
\[\frac{1 - \sin \theta}{\cos \theta}\] is equal to
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
Prove the following identities.
cot θ + tan θ = sec θ cosec θ
Prove that cot2θ × sec2θ = cot2θ + 1
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
