Advertisements
Advertisements
प्रश्न
`(1 + cot^2 theta ) sin^2 theta =1`
Advertisements
उत्तर
LHS= `(1+cot^2 theta)sin^2 theta`
=`cosec^2 theta sin^2 theta (∵ cosec^2 theta - cot^2 theta =1)`
=`1/(sin ^2theta)xxsin^2 theta`
=1
Hence, LHS = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
(sec A – cos A) (sec A + cos A) = sin2 A + tan2 A
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
`((sin A- sin B ))/(( cos A + cos B ))+ (( cos A - cos B ))/(( sinA + sin B ))=0`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Prove the following identities:
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that: `(sec θ - tan θ)/(sec θ + tan θ ) = 1 - 2 sec θ.tan θ + 2 tan^2θ`
Prove that: 2(sin6θ + cos6θ) - 3 ( sin4θ + cos4θ) + 1 = 0.
Prove that `(sin 70°)/(cos 20°) + (cosec 20°)/(sec 70°) - 2 cos 70° xx cosec 20°` = 0.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
If 1 – cos2θ = `1/4`, then θ = ?
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
