Question

Given $cot\theta =\frac{7}{8}$
$\frac{1+sin\theta }{cos\theta }$

Answer 1

$\cot \theta =\frac{\text{Adjacent side of}\theta }{\text{Opposite side of}\theta }=\frac{7}{8}$
Let $ABC$ is a right angle triangle
$AB=7$ and $BC=8$
By using Pythagoras theorem
$A{C}^2=A{B}^2+B{C}^2$
$=7^2+8^2=49+64=113$
$\therefore AC=\sqrt{113}$
$\therefore \sin \theta =\frac{\text{Opposite side of}\theta }{\text{Hypotenuse}}$
$=\frac{BC}{AC}=\frac{8}{\sqrt{113}}$
$\therefore cos\theta =\frac{\text{Adjacent side of}\theta }{\text{Hypotenuse}}$
$=\frac{AB}{AC}=\frac{7}{\sqrt{113}}$
$\frac{1+\sin \theta }{\cos \theta }=\frac{1+\frac{8}{\sqrt{113}}}{\frac{7}{\sqrt{113}}}=\frac{\sqrt{113}+8}{7}$