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प्रश्न
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
उत्तर
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
⇒ `sin(90^circ - 18^circ)/cos18^circ - sec(90^circ - 58^circ)/(cosec58^circ)`
⇒ `cos18^circ/cos18^circ - (cosec 58^circ)/(cosec58^circ) = 1 - 1 = 0`
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संबंधित प्रश्न
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
