Question:
Evaluate
\[
\tan x+\frac{\cos x}{1+\sin x}.
\]
Evaluate
\[
\tan x+\frac{\cos x}{1+\sin x}.
\]
Show Hint
Whenever an expression contains \(\sin^2x+\cos^2x\), replace it by \(1\). This often leads to immediate simplification.
Updated On: Jun 18, 2026
- \(\tan 2x\)
- \(\cosec x\)
- \(\sec x\)
- \(\cos 2x\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Express \(\tan x\) in terms of sine and cosine.
\[ \tan x=\frac{\sin x}{\cos x}. \] Therefore, \[ \tan x+\frac{\cos x}{1+\sin x} = \frac{\sin x}{\cos x} + \frac{\cos x}{1+\sin x}. \]
Step 2: Take the LCM.
\[ = \frac{\sin x(1+\sin x)+\cos^2x} {\cos x(1+\sin x)}. \] Expanding the numerator, \[ = \frac{\sin x+\sin^2x+\cos^2x} {\cos x(1+\sin x)}. \] Using \[ \sin^2x+\cos^2x=1, \] we get \[ = \frac{\sin x+1} {\cos x(1+\sin x)}. \]
Step 3: Simplify.
Cancelling \((1+\sin x)\), \[ = \frac{1}{\cos x}. \] Therefore, \[ = \sec x. \]
Step 4: Final conclusion.
Hence, \[ \boxed{\sec x} \]
\[ \tan x=\frac{\sin x}{\cos x}. \] Therefore, \[ \tan x+\frac{\cos x}{1+\sin x} = \frac{\sin x}{\cos x} + \frac{\cos x}{1+\sin x}. \]
Step 2: Take the LCM.
\[ = \frac{\sin x(1+\sin x)+\cos^2x} {\cos x(1+\sin x)}. \] Expanding the numerator, \[ = \frac{\sin x+\sin^2x+\cos^2x} {\cos x(1+\sin x)}. \] Using \[ \sin^2x+\cos^2x=1, \] we get \[ = \frac{\sin x+1} {\cos x(1+\sin x)}. \]
Step 3: Simplify.
Cancelling \((1+\sin x)\), \[ = \frac{1}{\cos x}. \] Therefore, \[ = \sec x. \]
Step 4: Final conclusion.
Hence, \[ \boxed{\sec x} \]
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