Question

answer this question fast​

Answer 1

$\bf{\underline{\underline{Question}}}$

Prove that

$\bf{\dfrac{Cos\theta}{Sin(90^\circ-\theta)} + \dfrac{Sin\theta}{Cos(90^\circ-\theta)} = 2}$

$\bf{\underline{\underline{Solution}}}$

$\bf{\underline{LHS}}$

= $\bf{\dfrac{Cos\theta}{Sin(90^\circ-\theta)} + \dfrac{Sin\theta}{Cos(90^\circ-\theta)}}$

= $\bf{\underline{We\:know}}$

$\bf{Sin(90^\circ-\theta) = Cos\theta}$

$\bf{Cos(90^\circ-\theta) = Sin\theta}$

$\bf{\underline{Hence}}$

= $\bf{\dfrac{Cos\theta}{Cos\theta} + \dfrac{Sin\theta}{Sin\theta}}$

=$\bf { \dfrac{ \not{\cancel{Cos\theta}} }{ \not{\cancel{Cos\theta}} }} + { \dfrac{ \not{\cancel{Sin\theta}}}{ \not{\cancel{Sin\theta}}}}$

= $\bf{1+1}$

= $\bf{2\: = \:(RHS)\:\:\:{\underline{\boxed{Proved}}}}$

$\bf{\underline{\underline{More\:Informations}}}$

• $\bf{Sin(90^\circ - \theta) = Cos\theta}$

• $\bf{Cos(90^\circ-\theta)=Sin\theta}$

• $\bf{Tan(90^\circ-\theta) = Cot\theta}$

• $\bf{Cot(90^\circ-\theta) = Tan\theta}$

• $\bf{Sec(90^\circ-\theta) = Cosec\theta}$

• $\bf{Cosec(90^\circ-\theta) = Sec\theta}$