Question

31. Prove that sin theta (1 + tan theta) + cos theta (1 + cot theta) = sec theta + cosec theta
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Answer 1

★To prove:-

• Sinθ(1 + Tanθ) + Cosθ(1 + Cotθ) = Secθ  + Cosecθ

★Proof:-

LHS:Sinθ(1 + Tanθ) + Cosθ(1 + Cotθ)......(1)

We know:

$\leadsto \underline{\boxed{\sf tan\theta = \dfrac{sin\theta }{cos\theta }}}$

$\leadsto \underline{\boxed{\sf cot\theta = \dfrac{cos\theta }{sin\theta } }}$

Putting these in equation (1),

$\displaystyle :\implies \sf sin\theta (1+\frac{sin\theta }{cos\theta} )+cos\theta (1+\frac{cos\theta }{sin\theta } )\\\\\\:\implies \sf sin\theta (\frac{cos\theta +sin\theta }{cos\theta} )+cos\theta (\frac{sin\theta +cos\theta }{sin\theta } )\\\\\\:\implies \sf \frac{sin\theta }{cos\theta } (sin\theta +cos\theta ) + \frac{cos\theta }{sin\theta} (sin\theta +cos\theta )\\\\\\:\implies \sf (sin\theta +cos\theta )(\frac{sin\theta }{cos\theta } + \frac{cos\theta }{sin\theta } )$

$\displaystyle :\implies \sf (sin\theta +cos\theta )(\frac{sin^2\theta +cos^2\theta )}{cos\theta sin\theta } )$

Using the formula,

$\leadsto \underline{\boxed{\sf sin^2\theta +cos^2\theta = 1}}$

$\displaystyle :\implies \sf (sin\theta +cos\theta )(\frac{1}{cos\theta sin\theta } )\\\\\\:\implies \sf \frac{cos\theta +sin\theta }{cos\theta sin\theta } \\\\\\:\implies \sf \frac{\cancel{cos\theta }}{\cancel{cos\theta }sin\theta } +\frac{\cancel{sin\theta }}{cos\theta \cancel{sin\theta }} \\\\\\:\implies \sf \frac{1}{sin\theta } +\frac{1}{cos\theta }$

Using the formula's:-

$\leadsto \underline{\boxed{\sf cosec\theta = \frac{1}{sin\theta } }}\\\\\leadsto \underline{\boxed{\sf sec\theta = \frac{1}{cos\theta } }}$

Therefore,

$\mapsto \boxed{\boxed{\sf cosec\theta + sec\theta }}$

Hence proved !

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