Advertisements
Advertisements
प्रश्न
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Advertisements
उत्तर
Given,
m = a sec A + b tan A and n = a tan A + b sec A
m2 – n2 = (a sec A + b tan A)2 – ( a tan A + b sec A)2
= a2 sec2 A + b2 tan2 A + 2ab sec A tan A – (a2 tan2 A + b2 sec2 A + 2ab sec A tan A)
= sec2 A (a2 – b2) + tan2 A (b2 – a2)
= (a2 – b2) [sec2 A – tan2 A]
= (a2 – b2) [Since sec2 A – tan2 A = 1]
Hence, m2 – n2 = a2 – b2
APPEARS IN
संबंधित प्रश्न
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove the following identity :
`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
