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प्रश्न
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
उत्तर
L.H.S = `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)`
= `1/("cosec" theta)(cottheta + tantheta)` .....`[(because tan(90 - theta) = cot theta),(cot(90 - theta) = tantheta)]`
= sin θ (cot θ + tan θ)
= `sintheta ((costheta)/(sintheta) + (sintheta)/(costheta))`
= `sintheta ((cos^2theta + sin^2theta)/(sintheta costheta))`
= `sintheta (1/(sintheta costheta))` ......[∵ sin2θ + cos2θ = 1]
= `1/costheta`
= sec θ
= R.H.S
∴ `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
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