Question

If $\tan \theta +\sec \theta =4$, then find $\sin \theta +\cos \theta$

• A. $1$
• B. $-1$
• C. $\frac{23}{17}$
• D. $\frac{13}{14}$

Answer 1

Answer: (C)

$\frac{23}{17}$

Given, $\tan \theta +\sec \theta =4$ ...(1)
But ${\sec }^2\theta -{\tan }^2\theta =1$
$\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1[\because \tan \theta +\sec \theta =4]$
$\Rightarrow 4(\sec \theta -\tan \theta )=1\Rightarrow (\sec \theta -\tan \theta )=\frac{1}{4}$ ...(2)
Solving (1) and (2), we get
$\Rightarrow 2\sec \theta =\frac{17}{4}$
$\Rightarrow \sec \theta =\frac{17}{8}$

$\Rightarrow \sec \theta =\frac{H}{B}=\frac{17}{8}=\frac{ML}{MN}$
$LN=\sqrt{{17}^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$
$\sin \theta =\frac{P}{H}=\frac{LN}{ML}=\frac{15}{17}$
$\cos \theta =\frac{B}{H}=\frac{MN}{ML}=\frac{8}{17}$
Required Expression
$\sin \theta +\cos \theta =\frac{15}{17}+\frac{8}{17}=\frac{15+8}{17}=\frac{23}{17}$