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प्रश्न
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
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उत्तर
LHS = `cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°)`
= `cos^2 26° + cos (90° - 26°) sin 26° + (tan 36°)/(cot (90° - 54°)`
= `cos^2 26° + sin 26°. sin 26° + (tan 36°)/(tan 36°)`
= cos2 26° + sin2 26 + 1 ....( cos2 θ + sin2 θ = 1)
= 1 + 1 = 2
= RHS
Hence proved.
संबंधित प्रश्न
Prove that (1 + cot θ – cosec θ)(1+ tan θ + sec θ) = 2
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(1 + cot A – cosec A)(1 + tan A + sec A) = 2
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`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
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`sqrt(cosec^2q - 1) = "cosq cosecq"`
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sec 60° = ?
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
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