Advertisements
Advertisements
प्रश्न
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
उत्तर
We have `(tan theta + sin theta ) = m and ( tan theta - sin theta )=n`
Now ,LHS = `(m^2-n^2)^2`
=`[(tan^2 theta + sin theta )^2 - "( tan theta - sin theta )^2]^2`
=`[(tan^2 theta + sin^2 theta + 2 tan theta sin theta )-( tan^2 theta + sin^2 theta -2 tan theta sin theta )]^2`
=`[(tan^2 theta +sin^2 theta + 2 tan theta sin theta - tan^2 theta - sin^2 theta+ 2 tan theta sin theta )]^2`
=`(4 tan theta sin theta )^2`
=`16 tan^2 theta sin^2 theta`
=`16 (sin ^2 theta )/(cos^2 theta ) sin^2 theta`
=`16 ((1- cos^2 theta) sin ^2 theta)/ cos^2 theta`
=` 16 [ tan^2 theta (1- cos^2 theta)]`
=`16 (tan^2 theta - tan^2 theta cos^2 theta)`
=`16 (tan^2 theta -(sin^2 theta)/(cos^2 theta) xx cos^2 theta )s`
=`16 ( tan^2 theta - sin^2 theta )`
=`16 (tan theta + sin theta ) ( tan theta - sin theta)`
=`16 mn [(tan theta + sin^theta )( tan theta - sin theta ) =mn]`
=`∴ (m^2 - n^2 )(m^2 - n^2 )^2 = 16 mn`
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) 2 (sin^6 θ + cos^6 θ) –3(sin^4 θ + cos^4 θ) + 1 = 0`
`(ii) (sin^8 θ – cos^8 θ) = (sin^2 θ – cos^2 θ) (1 – 2sin^2 θ cos^2 θ)`
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
Prove that cot2θ × sec2θ = cot2θ + 1
Prove the following:
`tanA/(1 + sec A) - tanA/(1 - sec A)` = 2cosec A
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
Prove the following trigonometry identity:
(sinθ + cosθ)(cosecθ – secθ) = cosecθ.secθ – 2 tanθ
