Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
उत्तर
We know that `sin^2 A + cos^2 A = 1`
So have
`sin^2 A cos^2 B - cos^2 A sin^2 B = sin^2 A (1 - sin^2 B) - (1 - sin^2 A) sin^2 B`
`= sin^2 A - sin^2 A sin^2 B - sin^2 B + sin^2 A sin^2 B`
`= sin^2 A - sin^2 B`
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
`(cot ^theta)/((cosec theta+1)) + ((cosec theta + 1))/cot theta = 2 sec theta`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If ` cot A= 4/3 and (A+ B) = 90° ` ,what is the value of tan B?
If 5x = sec ` theta and 5/x = tan theta , " find the value of 5 "( x^2 - 1/( x^2))`
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ