Advertisements
Advertisements
प्रश्न
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
उत्तर
`cos^2 26^@ + cos 64^@ sin 26^@ + tan 36^@/cot 54^@`
`= cos^2 26^@ + cos(90^@ - 26^@) sin 26^@ + tan 36^@/(cot(90^@ - 36^@))`
`= cos^2 26^@ + sin 26^@.sin26^@ + tan36^@/tan36^@` `[∵ cos(90^@ - theta) = sin theta, cot(90^@ - theta) = tan theta]`
`= cos^2 26^@ + sin^2 26^@ + 1`
`= 1 + 1 [∵ cos^2 theta + sin^2 theta = 1]`
= 2
APPEARS IN
संबंधित प्रश्न
`"If "\frac{\cos \alpha }{\cos \beta }=m\text{ and }\frac{\cos \alpha }{\sin \beta }=n " show that " (m^2 + n^2 ) cos^2 β = n^2`
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
Prove the following identities:
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove the following identity :
`(cosA + sinA)^2 + (cosA - sinA)^2 = 2`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2