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प्रश्न
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
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उत्तर
LHS = `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ))`
= `(sin θ. sin θ cos θ)/(cos θ) + (cos θ . cos θ sin θ)/(sin θ)`
= sin2 θ + cos2 θ
= 1
= RHS
Hence proved.
संबंधित प्रश्न
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
Write the value of cosec2 (90° − θ) − tan2 θ.
Prove the following identity :
`(1 + sinθ)/(cosecθ - cotθ) - (1 - sinθ)/(cosecθ + cotθ) = 2(1 + cotθ)`
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Choose the correct alternative:
Which is not correct formula?
Prove the following:
(sin α + cos α)(tan α + cot α) = sec α + cosec α
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
