Advertisements
Advertisements
प्रश्न
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
Advertisements
उत्तर
LHS= `(1+tan^2theta)(1+cot^2 theta)`
=`sec^2 theta. cosec^2 theta (∵ sec^2 theta - tan^2 theta=1 and cosec^2 - cot^2 theta =1)`
=`1/(cos^2 theta. sin^theta)`
=` 1/((1-sin^2 theta ) sin^2 theta`
=`1/(sin^2theta-sin^4theta)`
==RHS
Hence, LHS = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
`(sec^2 theta-1) cot ^2 theta=1`
`cosec theta (1+costheta)(cosectheta - cot theta )=1`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
Choose the correct alternative:
sec2θ – tan2θ =?
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
