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प्रश्न
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
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उत्तर
LHS = (sin θ + cos θ)(cosec θ – sec θ)
= `(sin θ + cos θ)(1/sin θ - 1/cos θ)`
= `(sin θ + cos θ)((cos θ - sin θ)/(sin θ * cos θ))`
= `(cos^2θ - sin^2θ)/(sinθ * cosθ)`
= `(1 - 2sin^2θ)/(sinθ*cosθ)`
= `1/(sinθ * cosθ) - (2 sin^2θ)/(sinθ * cosθ)`
= `cosec θ · sec θ - (2 sin^2 θ)/(sin θ * cos θ)`
= cosec θ · sec θ – 2 tan θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
`tan^2 theta - sin^2 theta tan^2 theta sin^2 theta`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Eliminate θ, if
x = 3 cosec θ + 4 cot θ
y = 4 cosec θ – 3 cot θ
Write the value of cosec2 (90° − θ) − tan2 θ.
Simplify
sin A `[[sinA -cosA],["cos A" " sinA"]] + cos A[[ cos A" sin A " ],[-sin A" cos A"]]`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove that : `1 - (cos^2 θ)/(1 + sin θ) = sin θ`.
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`
Which of the following is true for all values of θ (0° ≤ θ ≤ 90°)?
