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प्रश्न
Without using trigonometric identity , show that :
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
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उत्तर
`sec70^circ sin20^circ - cos20^circ cosec70^circ = 0`
Consider `sec70^circ sin20^circ - cos20^circ cosec70^circ`
⇒ `sec(90^circ - 20^circ)sin20^circ - cos20^circ . cosec(90^circ - 20^circ)`
⇒ `cosec20^circ sin20^circ - cos20^circ sec20^circ`
⇒ `1/sin20^circ . sin20^circ - cos20^circ . 1/cos20^circ`
⇒ 1 - 1 = 0
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संबंधित प्रश्न
Prove that: `sqrt((sec theta - 1)/(sec theta + 1)) + sqrt((sec theta + 1)/(sec theta - 1)) = 2 cosec theta`
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sec2A + cosec2A = sec2A . cosec2A
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
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Evaluate:
`(tan 65°)/(cot 25°)`
Prove that (cosec A - sin A)( sec A - cos A) sec2 A = tan A.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
Prove that `(sin^2theta)/(cos theta) + cos theta` = sec θ
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
