Advertisements
Advertisements
प्रश्न
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
Advertisements
उत्तर
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
consider `sin42^circ sec48^circ + cos42^circ cosec48^circ`
⇒ `sin42^circ sec(90^circ - 42^circ) + cos42^circ cosec(90^circ - 42^circ)`
⇒ `sin42^circ . cosec42^circ + cos42^circ sec42^circ`
⇒ `sin42^circ . 1/sin42^circ + cos42^circ 1/cos42^circ`
⇒ 1 + 1 = 2
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
Write True' or False' and justify your answer the following :
The value of sin θ+cos θ is always greater than 1 .
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
