Question

prove that sin theta (1 + tan theta) + cos theta (1 + cot theta) = sec theta + cosec theta

Answer 1

Answer :
Given
Sin $\theta$ ( 1 + tan $\theta$ ) + Cos $\theta$ ( 1 + Cot $\theta$ ) = Sec $\theta$ + Cosec $\theta$

Taking L.H.S.
$\Rightarrow$Sin $\theta$ ( 1 + tan $\theta$ ) + Cos $\theta$ ( 1 + Cot $\theta$ )

$\Rightarrow$Sin $\theta$ ( 1 + $\frac{Sin\theta }{Cos\theta }$ ) + Cos $\theta$ ( 1 + $\frac{Cos\theta }{Sin\theta }$ )

$\Rightarrow$Sin $\theta$ $\left(\frac{Cos\theta +\hspace{0.17em}Sin\theta }{Cos\theta }\right)$ + Cos $\theta$ $\left(\frac{Sin\theta +Cos\theta }{Sin\theta }\right)$

$\Rightarrow$$\frac{Sin\theta }{Cos\theta }$ ( Sin $\theta$ + Cos ​$\theta$ ) + $\frac{Cos\theta }{Sin\theta }$( Sin $\theta$ + Cos $\theta$ )

$\Rightarrow$( Sin $\theta$ + Cos ​$\theta$ ) ( $\frac{Sin\theta }{Cos\theta }$ + $\frac{Cos\theta }{Sin\theta }$ )

$\Rightarrow$( Sin $\theta$ + Cos ​$\theta$ ) $\left(\frac{Si{n}^2\theta +Co{s}^2\theta }{Sin\theta Cos\theta }\right)$

We know ( Sin ²$\theta$ + Cos²$\theta$ = 1 ) So, we get

$\Rightarrow$( Sin $\theta$ + Cos ​$\theta$ )$\left(\frac{1}{Sin\theta Cos\theta }\right)$

$\Rightarrow$$\left(\frac{Sin\theta }{Sin\theta Cos\theta }+\frac{Cos\theta }{Sin\theta Cos\theta }\right)$


$\Rightarrow$$\left(\frac{1}{Cos\theta }+\frac{1}{Sin\theta }\right)$

$\Rightarrow$Sec $\theta$ + Cosec $\theta$
Hence
L.H.S. = R.H.S. ( Hence proved )