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प्रश्न
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
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उत्तर
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80)`
`= (sin 35^@ . cos (90^@ - 35^@) + cos 35^@. sin (90^@ - 35^@))/(cosec^2(90^@ - 80^@) - tan^2 80^@`)
`= (sin 35^@ . sin 35^@ + cos 35^@ . cos 35^@) /(sec^2 80^@ - tan^2 80^@)`
`= (sin^2 35^@ + cos^2 35^@)/(sec^2 80^@ - tan^2 80^@) = 1/1 = 1`
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संबंधित प्रश्न
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sec2θ – tan2θ = ?
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Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
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