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प्रश्न
Prove the following:
`(sin^2(-160^circ))/(sin^(2)70^circ) + sin(180^circ - theta)/sintheta` = sec220°
उत्तर
sin(– 160°) = – sin 160°
= – sin(180° – 20°)
= – sin 20°
sin 70° = sin (90° – 20°) = cos 20°
∴ L.H.S. = `(sin^2(-160^circ))/(sin^(2)70^circ) + sin(180^circ - theta)/sintheta`
= `((-sin20^circ)^2)/((cos20^circ)^2) + sintheta/sintheta`
= `(sin^(2)20^circ)/(cos^(2)20^circ) + 1`
= 1 + tan220°
= sec220°
= R.H.S.
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