Advertisements
Advertisements
प्रश्न
If A = 60°, B = 30° verify that tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`.
Advertisements
उत्तर
It is given that A = 60°, B = 30°
Putting A = 60° and B = 30° in the given equation,
we get
tan( A - B) = `(tan A - tan B)/(1 + tan A. tan B)`
⇒ tan( 60° - 30° ) = `(tan 60° - tan 30° )/(1 + tan 60°. tan 30° )`
⇒ tan 30° = `(sqrt3 - 1/sqrt3)/(1 + sqrt3 xx 1/sqrt3)`
⇒ `((3-1)/sqrt3)/2`
⇒ `(2/sqrt3)/(2/1)`
⇒ `2/(2sqrt3)`
⇒ `1/sqrt3`
⇒ LHS = RHS
संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following identities:
`tan A - cot A = (1 - 2cos^2A)/(sin A cos A)`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
If `sec theta + tan theta = x," find the value of " sec theta`
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
If tan θ × A = sin θ, then A = ?
