Question

How do you simplify $\dfrac{1+\tan \theta }{1+\cot \theta }?$

Answer 1

Hint: We can solve this trigonometry question using different fundamental trigonometric identities. When it comes down to simplify with these identities. We must use combinations of these identities to reduce a much more complex expression to its simplest form. We will reciprocal identities for simplifying this function identities also.

Formula used:
$\sin \theta =\dfrac{1}{\cos \theta },\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cos \theta =\dfrac{\cos \theta }{\sin \theta },\dfrac{1}{\sin \theta }=\cos \theta $

Complete step by step solution:
Simplify = $\dfrac{1+\tan \theta }{1+\cot \theta }$
If you are not sure how to start then chars everything to sine cosine by using quotient identity. We
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
Put this in the equation we will get,
$\dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{1+\left( \dfrac{\sin \theta }{\cos \theta } \right)}{1+\left( \dfrac{\cos \theta }{\sin \theta } \right)}$
Multiply out:
$\dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{\dfrac{\cos \theta +\sin \theta }{\cos \theta }}{\dfrac{\sin \theta +\cos \theta }{\sin \theta }}$$\dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{\dfrac{1}{\cos x}}{\dfrac{1}{\sin x}}$
Now, apply the identity $\dfrac{1}{\cos \theta }=\sin \theta $and $\dfrac{1}{\sin x}=\cos x$
By applying the reciprocal identity we have,
$\dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{\sin \theta }{\cos \theta }$
As per quotient identity we have $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $
$\Rightarrow \dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{\sin \theta }{\cos \theta }$
As per quotient identity we have $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $
So, will get $\dfrac{1+\tan \theta }{1+\cot \theta }=\tan \theta $
So often the simplification $\dfrac{1+\tan \theta }{1+\cot \theta }=\tan \theta $
We have simplified this using various fundamental identities of trigonometry.

Additional Information:
We can simplify this by using only reciprocal identities. This is the another method of solving or simplifying this $\dfrac{1+\tan \theta }{1+\cot \theta }$
$=\dfrac{1+\tan \theta }{1+\cot \theta }$ Apply the reciprocal identity $\cot \theta =\dfrac{1}{\tan \theta }$
$\dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{1+\tan \theta }{1+\dfrac{1}{\tan \theta }}$
Multiply out:
$\dfrac{1+\tan \theta }{1+\cot \theta }=\dfrac{1+\tan \theta }{\dfrac{\tan \theta +1}{\tan \theta }}$
$\Rightarrow \dfrac{1+\tan \theta }{1+\cot \theta }=\tan \theta \left( \dfrac{1+\tan \theta }{\tan \theta +1} \right)$
Cancelling out $\tan \theta $ it appears both in the numerator and in the denominator.
$\dfrac{1+\tan \theta }{1+\cot \theta }=\tan \theta $
By solving it through another we get the same answer.
We can simplify any question about the use of identity. We will take another problem for your better understanding for ex.
Simplify $=\dfrac{\sin x}{\cos x}+\dfrac{\cos \theta }{1+\sin x}$
Put on a common denominator.
$=\dfrac{\sin x\left( 1+\sin x \right)}{\cos x\left( 1+\sin x \right)}+\dfrac{\cos \left( \cos x \right)}{\cos x\left( 1+\sin x \right)}$
Now, multiply out:
$\Rightarrow \dfrac{\sin x+{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos \left( 1+\sin x \right)}$
Apply the Pythagorean identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
$\Rightarrow \dfrac{\sin x+1}{\cos \left( 1+\sin x \right)}$
Cancelling out the $\sin x+1$ since it appears both in the numerator and in the denominator.
$\Rightarrow \dfrac{\sin x+1}{\cos x\left( 1+\sin x \right)}$
$\Rightarrow \dfrac{1}{\cos x}$
Applying the reciprocal identity $\dfrac{1}{\cos \theta }=\sec \theta $
$\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=\sec x$
After simplifying we have seen.

Note: When it comes down to simplifying with these identities. We must use a combination of these identities to reduce a much more complex expression to its simplest form. Memorise these identities because they are that to mathematics. I would highly recommend memorization.