Advertisements
Advertisements
प्रश्न
Prove the following trigonometric identities.
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Advertisements
उत्तर
In the given question, we need to prove
`(1/(sec^2 theta - cos theta) + 1/(cosec^2 theta - sin^2 theta)) sin^2 theta cos^2 theta = (1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta + cos^2 theta)`
Now using `sec theta = 1/ cos theta` and `cosec theta = 1/sin theta` in LHS we get
LHS =`(1/((1/cos^2 theta) - cos^2 theta) + 1/((1/sin^2 theta) - sin^2 theta)) sin^2 theta cos^2 theta`
`= (1/((1 - cos^4 theta)/cos^2 theta) + 1/((1 - sin^4 theta)/sin^2 theta)) sin^2 theta cos^2 theta`
`= ((cos^2 theta)/(1 - cos^4 theta) + sin^2 theta/(1 - sin^4 theta)) sin^2 theta cos^2 theta`
Further using the identity `a^2 - b^2 = (a + b)(a- b)` we get
LHS = `(cos^2 theta/((1 - cos^2 theta)(1 + cos^2 theta)) + sin^2 theta/((1 - sin^2 theta) (1 + sin^2 theta)))sin^2 theta cos^2 theta`
`= ((cos^2 theta)/(sin^2 theta(1 + cos^2 theta)) + sin^2 theta/(cos^2 theta(1 + sin^2 theta))) sin^2 theta cos^2 theta`
`= ((cos^2 theta(cos^2 theta(1 + sin^2 theta))+sin^2 theta(sin^2 theta(1 + cos^2 theta)))/(sin^2 theta cos^2 theta(1 + cos^2 theta)(1 +sin^2 theta))) sin^2 theta cos^2 theta`
`= ((cos^4 theta(1 + sin^2 theta) + sin^4 theta(1 + cos^2 theta))/((1 + cos^2 theta)(1 + sin^2 theta)))`
Further using the identity `sin^2 theta + cos^2 theta = 1` we get
LHS = `((cos^4 theta + cos^4 theta sin^2 theta + sin^4 theta + sin^4 theta cos^2 theta)/(1 + cos^2 theta + sin^2 theta + sin^2 theta cos^2 theta))`
`= (cos^4 theta + sin^4 theta + cos^2 theta sin^2 theta (cos^2 theta + sin^2 theta)) /(2 + sin^2 theta cos^2theta)`
`= ((cos^4 theta +sin^4 theta +cos^2 theta sin^2theta (1))/(2 + sin^2 theta cos^2 theta))`
Now, from the identity `a^2 + b^2 = (a + b)^2 - 2ab` we get
So,
LHS = `(((cos^2 theta + sin^2 theta)^2 - 2cos^2 theta sin^2 theta +cos^2 theta sin^2 theta)/(2 + sin^2 theta cos^2 theta))`
`= (((1)^2 - cos^2 theta sin^2 theta)/(22 +sin^2 theta cos^2 theta))`
`= ((1 - sin^2 theta cos^2 theta)/(2 + sin^2 theta cos^2 theta))`
Hence proved.
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)`
Prove that (cosec A – sin A)(sec A – cos A) sec2 A = tan A.
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities.
`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Prove the following identities:
`cosecA - cotA = sinA/(1 + cosA)`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
Write the value of tan10° tan 20° tan 70° tan 80° .
Simplify : 2 sin30 + 3 tan45.
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`cosecA + cotA = 1/(cosecA - cotA)`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt(a^2 + b^2 - c^2)`
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ.
