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प्रश्न
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
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उत्तर
LHS = `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A)`
= `((sin A + cos A)^2 + (sin A - cos A)^2)/((sin A - cos A)(sin A + cos A))`
= `(sin^2 A + cos^2 A + 2 sin A cos A + sin^2 A + cos^2 A - 2sin A. cos A)/(sin^2 A - cos^2 A)`
= `2(sin^2 A + cos^2 A)/(sin^A - cos^2 A)`
= `2/(sin^2 A - cos^2 A)`
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
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Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
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Show that one of the values of each member of this equality is sin α sin β sin γ
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Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
If `cos theta = 2/3 , " write the value of" (4+4 tan^2 theta).`
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
Choose the correct alternative:
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