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प्रश्न
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
उत्तर
LHS = `(sin theta - cos theta )/ (sin theta + cos theta) +( sin theta + cos theta )/( sin theta - cos theta )`
=` ((sin theta - cos theta )^2 + (( sin theta + cos theta )^2))/((sin theta + cos theta )( sin theta - cos theta ))`
=` (sin^2 theta + cos ^2 theta -2 sin theta cos theta + sin^2 theta + cos^2 theta + 2 sin theta cos theta)/( sin^ 2theta - cos^ 2theta)`
=` (1+1)/(sin^2 theta - ( 1-sin ^2 theta)) ( ∵ sin^2 theta + cos^2 theta =1)`
=`2/(sin^2 theta - 1 + sin^2 theta)`
=` 2/ (sin^2 theta -1)`
= RHS
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संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
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`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
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Prove the following trigonometric identities.
`tan theta + 1/tan theta = sec theta cosec theta`
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
(1 + cot A − cosec A) (1 + tan A + sec A) = 2
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`(1 - cos^2θ)sec^2θ = tan^2θ`
Find the value of ( sin2 33° + sin2 57°).
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Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or `1/2`.
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
