Advertisements
Advertisements
प्रश्न
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
Advertisements
उत्तर
LHS =` 1+(cot^2 theta)/((1+ cosectheta))`
=`1+((cosec^2 theta-1))/((cosectheta++1)) (∵ cosec^2 theta - cot^2 theta =1)`
=`1+((cosectheta+1)(cosec theta-1))/((cosec theta +1))`
=`1+ (cosec theta -1)`
=` cosec theta`
=RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`
`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`
Prove the following trigonometric identity.
`cos^2 A + 1/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
Prove the following identities:
`(1 + (secA - tanA)^2)/(cosecA(secA - tanA)) = 2tanA`
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
For ΔABC , prove that :
`sin((A + B)/2) = cos"C/2`
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
Show that tan 7° × tan 23° × tan 60° × tan 67° × tan 83° = `sqrt(3)`.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
