Advertisements
Advertisements
प्रश्न
Prove that `sqrt(sec^2 theta + cosec^2 theta) = tan theta + cot theta`
उत्तर
L.H.S = `sqrt(sec^2 theta + cosec^2 theta)`
`= sqrt(1/(cos^2 theta) + 1/(sin^2 theta))`
`= sqrt((sin^2 theta + cos^2 theta)/(cos^2 theta sin ^2 theta))`
`= sqrt(1/(cos^2 theta sin^2 theta))`
`= sqrt(1/(cos^2 theta) xx 1/(sin^2 theta))`
`= sqrt(sec^2 theta xx cosec^2 theta)`
`=sec theta xx cosec theta`
R.H.S = `tan theta + cot theta = (sin theta)/(cos theta)+ cos theta/sin theta = (sin^2 theta + cos^2 theta)/(cos theta sin theta) = 1/cos theta xx 1/sin theta = sec theta xx cosec theta`
Thus L.H.S = R.H.S
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = sec A + tan A`
Show that : tan 10° tan 15° tan 75° tan 80° = 1
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
What is the value of (1 + cot2 θ) sin2 θ?
Choose the correct alternative:
1 + tan2 θ = ?
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
Choose the correct alternative:
cos 45° = ?