Advertisements
Advertisements
प्रश्न
`costheta/((1-tan theta))+sin^2theta/((cos theta-sintheta))=(cos theta+ sin theta)`
Advertisements
उत्तर
LHS = `cos theta/((1-tan theta))-sin^2theta/((cos theta-sintheta))`
=`cos theta/((1-sintheta/costheta)) -sin^2 theta/((cos theta-sin theta))`
=`cos^2 theta/((cos theta-sintheta))- sin^2 theta/((cos theta-sin theta))`
=`(cos^2 theta- sin ^2 theta)/((cos theta- sin theta))`
=`((costheta + sin theta)( cos theta-sin theta))/((cos theta - sin theta))`
=`(cos theta + sin theta)`
= RHS
Hence, LHS = RHS
APPEARS IN
संबंधित प्रश्न
If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`
9 sec2 A − 9 tan2 A = ______.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
`sin^2 theta + cos^4 theta = cos^2 theta + sin^4 theta`
`(cos^3 theta +sin^3 theta)/(cos theta + sin theta) + (cos ^3 theta - sin^3 theta)/(cos theta - sin theta) = 2`
If `secθ = 25/7 ` then find tanθ.
Write True' or False' and justify your answer the following :
The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Evaluate:
`(tan 65^circ)/(cot 25^circ)`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
Prove that sin4A – cos4A = 1 – 2cos2A
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
