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प्रश्न
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
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उत्तर
L.H.S. = cosec A(1 + cos A) (cosecA – cot A)
= `1/(sin A)(1 + cos A)(1/(sin A) - (cos A)/(sin A))`
`((1-cos A)/sin A)`
`1/sin A(1+cos A)((1-cos A)/sin A)`
`= ((1+ cos A)(1-cos A))/sin^2 A`
Apply the identity (1 + cosA) (1 − cosA) = 1 − cos2A
`= (1-cos^2A)/sin^2A`
`= sin^2A/sin^2A = 1`
cscA(1 + cosA) (cscA − cotA) = 1 proved
संबंधित प्रश्न
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
Show that none of the following is an identity:
`cos^2theta + cos theta = 1`
Write 'True' or 'False' and justify your answer in the following:
The value of \[\sin \theta\] is \[x + \frac{1}{x}\], where 'x' is a positive real number.
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
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)
