Advertisements
Advertisements
Question
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Advertisements
Solution
LHS = `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2`
= `((sin 20°. sin 70°)/(cos 20°))^2 + ((cos 20°. cos 20°)/(sin 20°))^2`
= `[(sin 20°.sin (90° - 20°))/(cos 20°)]^2 + [(cos 20°. cos(90° - 20°))/(sin 20°)]^2`
= `[ (sin 20°.cos 20°)/(cos 20°)]^2 + [(cos 20°. sin 20°)/(sin 20°)]^2`
= sin2 20° + cos2 20°
= 1
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
Prove that:
`"tan A"/(1 + "tan"^2 "A")^2 + "Cot A"/(1 + "Cot"^2 "A")^2 = "sin A cos A"`.
9 sec2 A − 9 tan2 A is equal to
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
