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Prove That: Sin 70 ∘ Cos 20 ∘ + Cosec 20 ∘ Sec 70 ∘ − 2 Cos 70 ∘ Cosec 20 ∘ = 0 - Mathematics

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प्रश्न

Prove that:

`(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2  cos 70^circ "cosec"  20^circ = 0`

योग

उत्तर

LHS = `(sin 70^circ)/(cos 20^circ) + ("cosec" 20^circ)/(sec 70^circ) - 2  cos 70^circ "cosec"  20^circ`

`=("sin"70^circ)/(sin(90^circ - 20^circ)) + (sec(90^circ-20^circ))/sec 70^circ - 2 cos 70^circ sec(90^circ-20^circ)` 

`= 1 + 1 - 2 xxcos 70^circxx1/cos 70^circ`

= 2 - 2

= 0

= RHS

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अध्याय 7: Trigonometric Ratios of Complementary Angles - Exercises [पृष्ठ ३१३]

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आरएस अग्रवाल Mathematics [English] Class 10
अध्याय 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 4.1 | पृष्ठ ३१३

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