Advertisements
Advertisements
प्रश्न
Evaluate:
`(tan 65°)/(cot 25°)`
Advertisements
उत्तर
`(tan 65°)/(cot 25°)`
= `(tan 90° - 25°)/(cot 25°)` ...(∵ tan(90°−θ) = cotθ )
= `( cot 25° )/( cot 25°)`
= 1
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`
`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`
The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ
