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प्रश्न
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
उत्तर
In the given question, we need to prove `1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Here, we will first solve the L.H.S.
Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get
`1/(sec A + tan A) - 1/cos A = 1/(1/cos A + sin A/cos A) - (1/cos A)`
`= 1/(((1 + sin A)/cos A)) - (1/cos A)`
`= (cos A/(1 + sin A)) - (1/cos A)`
`= (cos^2 A - (1 + sin A))/((1 + sin A)(cos A))`
On further solving, we get
`(cos^2 A -(1 + sin A))/((1 + sin A)(cos A)) = (cos^2 A - 1 - sin A)/((1 + sin A)(cos A))`
`= (-sin^2 A - sin A)/((1 + sin A)(cos A))` (Using `sin^2 theta = 1 - cos^2 theta)`
`= (-sin A(sin A + 1))/((1 + sin A)(cos A))`
`= (-sin A)/cos A`
= -tan A
Similarly we solve the R.H.S.
`((1 - sin A) - cos^2 A)/((cos A)(1 - sin^2 A)) = (1 - sin A - cos^2 A)/((cos A)(1 - sin A))`
`= (sin^2 A - sin A)/((cos A)(1 - sin A))` (Using `sin^2 theta = 1- cos^2 theta`)
`= (-sin A(1 - sin A))/((cos A)(1 - sin A))`
`= (-sin A)/cos A`
= - tan A
So, L.H.S = R.H.S
Hence proved.
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= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
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= R.H.S
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