Question

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

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$\mathsf{Prove\: that\: sin\, \theta(1+tan\, \theta)+cos\, \theta(1+cot\, \theta) = sec\, \theta+cos\, \theta}$




$\underline{\underline{\Huge{\textbf{Solution:}}}}$

$\underline{\LARGE{\textsf{Step-1:}}}$
$\textsf{Consider LHS}$

$\underline{\Large{\textbf{LHS = }}}$

$\implies \mathsf{sin\, \theta(1+tan\, \theta)+cos\, \theta(1+cot\, \theta)}$




$\underline{\LARGE{\textsf{Step-2:}}}$
$\textsf{Express tan and cot in terms of}\\\textsf{sin and cos and simplify}$

$\LARGE{\boxed{\mathbf{\begin{aligned}\quad\bigstar\; \; tan\, \theta &= \dfrac{sin\, \theta}{cos\, \theta}\\\\\quad \bigstar\; \; \cot\, \theta &= \dfrac{cos\, \theta}{sin\, \theta}\quad\end{aligned}}}}$

$\implies \mathsf{sin\, \theta\left(1+\dfrac{sin\, \theta}{cos\, \theta}\right)+cos\, \theta\left(1+\dfrac{cos\, \theta}{sin\, \theta}\right)}\\\\\implies \mathsf{sin\, \theta\left(\dfrac{cos\, \theta + sin\, \theta}{cos\, \theta}\right)+cos\, \theta\left(\dfrac{sin\, \theta+ cos\, \theta}{sin\, \theta}\right)}\\\\\implies \mathsf{\dfrac{sin\, \theta}{cos\, \theta}(sin\, \theta+cos\, \theta)+\dfrac{cos\, \theta}{sin\, \theta}\left(sin\, \theta+ cos\, \theta\right)}$




$\underline{\LARGE{\textsf{Step-3:}}}$
$\textsf{Factor (sin\, \theta +cos\, \theta) from the two terms and simplify}$

$\implies \mathsf{(sin\, \theta+cos\, \theta)\, \left(\dfrac{sin\, \theta}{cos\, \theta}+ \dfrac{cos\, \theta}{sin\, \theta}\right)}\\\\\implies \mathsf{(sin\, \theta+cos\, \theta)\, \left(\dfrac{sin^2\, \theta+cos^\, \theta}{cos\, \theta\: sin\, \theta}\right)}$


$\textsf{Using the Identity}$
$\LARGE{\boxed{\quad\bigstar\; \; \mathbf{sin^2\, \theta+cos^2\, \theta = 1}\quad}}$


$\implies \mathsf{(sin\, \theta+cos\, \theta)\, \left(\dfrac{1}{cos\, \theta\: sin\, \theta}\right)}\\\\\implies \mathsf{\dfrac{sin\, \theta+cos\, \theta}{cos\, \theta\: sin\, \theta}}\\\\\implies \mathsf{\dfrac{\cancel{sin\, \theta}}{cos\, \theta\: \cancel{sin\, \theta}}+\dfrac{\cancel{cos\, \theta}}{\cancel{cos\, \theta}\: sin\, \theta}}\\\\\implies \mathsf{\dfrac{1}{cos\, \theta}+\dfrac{1}{sin\, \theta}}$

$\textsf{we have}$
$\LARGE{\boxed{\mathbf{\begin{aligned}\quad\bigstar\; \; sec\, \theta &= \dfrac{1}{cos\, \theta}\\\\\quad \bigstar\; \; cosec\, \theta &= \dfrac{1}{sin\, \theta}\quad\end{aligned}}}}$

$\implies \mathsf{sec\, \theta+cosec\, \theta}$

$\underline{\Large{\textbf{= RHS}}}$


$\underline{\LARGE{\textbf{Hence Proved}}}$