How do you simplify \[\tan A\sin A+\cos A\]?

Hint: Assume the given expression as ‘E’. Use the conversion formula: - \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] while leaving the other terms as it is. Now, take the L.C.M. and simplify the expression. Use the trigonometric identity: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and finally use the conversion: - \[\dfrac{1}{\cos \theta }=\sec \theta \] to get the answer.

Complete step by step solution:
Here, we have been provided with the trigonometric expression: \[\tan A\sin A+\cos A\] and we are asked to simplify it. Here, we are going to use some basic trigonometric identity to simplify this expression. Let us assume this expression as ‘E’, so we have,
\[\Rightarrow E=\tan A\sin A+\cos A\]
We know that the tangent function is the ratio of sine function and cosine function, mathematically given as: - \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \], so using this conversion we have,
\[\begin{align}
  & \Rightarrow E=\dfrac{\sin A}{\cos A}\times \sin A+\cos A \\
 & \Rightarrow E=\dfrac{{{\sin }^{2}}A}{\cos A}+\cos A \\
\end{align}\]
Taking L.C.M. in the R.H.S., we get,
\[\Rightarrow E=\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A}\]
We know that in any right-angle triangle we have \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. So, applying this identity in the above expression ‘E’, we have,
\[\Rightarrow E=\dfrac{1}{\cos A}\]
We know that the reciprocal of cosine function is called as the secant function, so we have \[\dfrac{1}{\cos \theta }=\sec \theta \],
\[\Rightarrow E=\sec A\]
Hence, the simplified value of the given expression is \[\sec A\] which is our answer.

Note:
You must remember the three basic trigonometric identities given as: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], \[1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta \] and \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], as they are used everywhere. There can be a different approach also using which you can solve the question. What we can do is, we will take the cosine function common from both the terms such that the equation will become \[E=\cos A\left( {{\tan }^{2}}A+1 \right)\]. Now, using the identity \[\left( 1+{{\tan }^{2}}\theta \right)={{\sec }^{2}}\theta \] we will get \[E=\cos A\times {{\sec }^{2}}\theta \]. Finally, using the formula \[\sec \theta \times \cos \theta =1\], we will get the value of expression ‘E’ equal to \[\sec A\]. We cannot simplify the expression any further. So, \[\sec A\] will be our answer.