Question:
Evaluate
\[
\int \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)}\,dx
\]
Evaluate
\[
\int \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)}\,dx
\]
Show Hint
For integrals involving \(\tan x\) and \(\tan(x+a)\), convert the expression into sine and cosine form, then use trigonometric identities to simplify.
Updated On: Jun 22, 2026
- \(\tan a\left(\log(\sec(x+a))+\log \sec x+C\right)\)
- \(\cot a\left(\log|\sin x|-\log|\sin(x+a)|\right)+C\)
- \(\tan a\left(\log\left(\dfrac{\cos x}{\sin(x+a)}\right)\right)+C\)
- \(\cot a\left(\log\dfrac{\sin(x+a)}{\cos(x+a)}\right)+C\)
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The Correct Option is B
Solution and Explanation
Step 1: Simplify the integrand.
Given integral is \[ \int \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)}\,dx \] Separating the terms, \[ \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)} = \frac{1}{\tan x\tan(x+a)}+1 \] \[ = \cot x\cot(x+a)+1 \]
Step 2: Convert into sine and cosine form.
\[ \cot x\cot(x+a)+1 = \frac{\cos x\cos(x+a)}{\sin x\sin(x+a)}+1 \] \[ = \frac{\cos x\cos(x+a)+\sin x\sin(x+a)}{\sin x\sin(x+a)} \] Using identity, \[ \cos A\cos B+\sin A\sin B=\cos(A-B) \] we get \[ \cos x\cos(x+a)+\sin x\sin(x+a)=\cos a \] Therefore, \[ \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)} = \frac{\cos a}{\sin x\sin(x+a)} \]
Step 3: Use cotangent difference identity.
Now, \[ \cot x-\cot(x+a) = \frac{\cos x}{\sin x}-\frac{\cos(x+a)}{\sin(x+a)} \] \[ = \frac{\cos x\sin(x+a)-\sin x\cos(x+a)}{\sin x\sin(x+a)} \] Using identity, \[ \sin B\cos A-\cos B\sin A=\sin(B-A) \] we get \[ \cos x\sin(x+a)-\sin x\cos(x+a)=\sin a \] Thus, \[ \cot x-\cot(x+a)=\frac{\sin a}{\sin x\sin(x+a)} \] So, \[ \frac{1}{\sin x\sin(x+a)} = \frac{\cot x-\cot(x+a)}{\sin a} \] Hence, \[ \frac{\cos a}{\sin x\sin(x+a)} = \frac{\cos a}{\sin a}\left(\cot x-\cot(x+a)\right) \] \[ = \cot a\left(\cot x-\cot(x+a)\right) \]
Step 4: Integrate.
Therefore, \[ \int \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)}\,dx = \cot a\int \left(\cot x-\cot(x+a)\right)\,dx \] Now, \[ \int \cot x\,dx=\log|\sin x| \] and \[ \int \cot(x+a)\,dx=\log|\sin(x+a)| \] So, \[ = \cot a\left(\log|\sin x|-\log|\sin(x+a)|\right)+C \]
Step 5: Final conclusion.
Hence, \[ \boxed{\cot a\left(\log|\sin x|-\log|\sin(x+a)|\right)+C} \]
Given integral is \[ \int \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)}\,dx \] Separating the terms, \[ \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)} = \frac{1}{\tan x\tan(x+a)}+1 \] \[ = \cot x\cot(x+a)+1 \]
Step 2: Convert into sine and cosine form.
\[ \cot x\cot(x+a)+1 = \frac{\cos x\cos(x+a)}{\sin x\sin(x+a)}+1 \] \[ = \frac{\cos x\cos(x+a)+\sin x\sin(x+a)}{\sin x\sin(x+a)} \] Using identity, \[ \cos A\cos B+\sin A\sin B=\cos(A-B) \] we get \[ \cos x\cos(x+a)+\sin x\sin(x+a)=\cos a \] Therefore, \[ \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)} = \frac{\cos a}{\sin x\sin(x+a)} \]
Step 3: Use cotangent difference identity.
Now, \[ \cot x-\cot(x+a) = \frac{\cos x}{\sin x}-\frac{\cos(x+a)}{\sin(x+a)} \] \[ = \frac{\cos x\sin(x+a)-\sin x\cos(x+a)}{\sin x\sin(x+a)} \] Using identity, \[ \sin B\cos A-\cos B\sin A=\sin(B-A) \] we get \[ \cos x\sin(x+a)-\sin x\cos(x+a)=\sin a \] Thus, \[ \cot x-\cot(x+a)=\frac{\sin a}{\sin x\sin(x+a)} \] So, \[ \frac{1}{\sin x\sin(x+a)} = \frac{\cot x-\cot(x+a)}{\sin a} \] Hence, \[ \frac{\cos a}{\sin x\sin(x+a)} = \frac{\cos a}{\sin a}\left(\cot x-\cot(x+a)\right) \] \[ = \cot a\left(\cot x-\cot(x+a)\right) \]
Step 4: Integrate.
Therefore, \[ \int \frac{1+\tan x\tan(x+a)}{\tan x\tan(x+a)}\,dx = \cot a\int \left(\cot x-\cot(x+a)\right)\,dx \] Now, \[ \int \cot x\,dx=\log|\sin x| \] and \[ \int \cot(x+a)\,dx=\log|\sin(x+a)| \] So, \[ = \cot a\left(\log|\sin x|-\log|\sin(x+a)|\right)+C \]
Step 5: Final conclusion.
Hence, \[ \boxed{\cot a\left(\log|\sin x|-\log|\sin(x+a)|\right)+C} \]
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