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प्रश्न
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
उत्तर
L.H.S = tan θ + tan (90° – θ) ...[∵ tan (90° – θ) = cot θ]
= tan θ + cot θ
= `sinθ/cosθ + cosθ/sinθ`
= `(sin^2 theta + cos^2 theta)/(sin theta cos theta)` ...`[∵ tan theta = sintheta/costheta "and" cot theta = costheta/sintheta]`
= `1/(sin theta cos theta)` ...[∵ sin2θ + cos2θ = 1]
= sec θ cosec θ ...`[∵ sec theta = 1/costheta "and" cos theta = 1/sin theta]`
= sec θ sec (90° – θ) ...[∵ sec (90° – θ) = cosec θ]
= R.H.S
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