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प्रश्न
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
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उत्तर
Solving the function using trignometric identities:
As we have `(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
LHS = `(sin theta - cos theta + 1)/(sin theta + cos theta - 1)`
Dividing the numerator and denomenator by cos θ
`(sin theta/cos theta - cos theta/cos theta + 1/cos theta)/(sin theta/cos theta + cos theta/cos theta - 1/cos theta)`
= `(tan theta - 1 + sec theta)/(tan theta + 1 - sec theta)`
Multiplying and dividing by (tan θ - sec θ),
= `(tan theta + sec theta - 1)/(tan theta - sec theta + 1)xx (tan theta - sec theta)/(tan theta - sec theta)`
[(tan θ + sec θ)(tan θ - sec θ = tan2θ - sec2θ)]
= `[((tan^2 theta - sec^2 theta) - (tan theta - sec theta))/((tan theta - sec theta + 1)(tan theta - sec theta))]`
Using the identity sec2θ - tan2 θ = 1,
= `((-1 - tan theta + sec theta))/([(tan theta - sec theta + 1)(tan theta - sec theta)])`
= `(-1)/(tan theta - sec theta)`
= `1/(sec theta - tan theta)`
= RHS Hence proved
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