# Evaluate $\sin{x}\tan{x}$ $+$ $\cos{x}$

The cosine of angle $x$ is added to the product of the sine of angle $x$ and tangent of angle $x$ in the given trigonometric expression, and the sine of $x$ times tan of $x$ plus cos of $x$ should be evaluated in this trigonometry problem. Let’s learn how to simplify the given trigonometric expression to find the value of sine of angle $x$ times tan of angle $x$ plus cos of angle $x$ by using the trigonometric identities.

### Express the Tan in terms of sine and cosine

There is a sine function in the first term and there is also a cosine function in the second term. So, it is a smart idea to convert the tan function in terms of sine and cosine, and it is possible by the quotient identity of sine and cosine functions.

$=\,\,$ $\sin{x}\Big(\dfrac{\sin{x}}{\cos{x}}\Big)$ $+$ $\cos{x}$

$=\,\,$ $\sin{x} \times \Big(\dfrac{\sin{x}}{\cos{x}}\Big)$ $+$ $\cos{x}$

### Find the Product by multiplying the factors

The second factor in the first term is a fraction and the multiplication symbol between them expresses that the product of them should be evaluated. So, let’s write the first factor in fraction form.

$=\,\,$ $\dfrac{\sin{x}}{1} \times \Big(\dfrac{\sin{x}}{\cos{x}}\Big)$ $+$ $\cos{x}$

Now, multiply the two fractions in the first term of the trigonometric expression to find their product.

$=\,\,$ $\dfrac{\sin{x} \times \sin{x}}{1 \times \cos{x}}$ $+$ $\cos{x}$

$=\,\,$ $\dfrac{\sin^2{x}}{\cos{x}}$ $+$ $\cos{x}$

### Find the sum of the two terms by the addition

The first term is a fraction but the second term is not a fraction. The addition of them can be performed only when the second term is in fraction form. So, let’s express the second term in fraction form.

$=\,\,$ $\dfrac{\sin^2{x}}{\cos{x}}$ $+$ $\dfrac{\cos{x}}{1}$

The denominators of the fractions in the trigonometric expression are different. So, the two unlike fractions cannot be added mathematically but it is possible when the fractions have same denominators. So, let’s make the denominator of the second term is same as the denominator of the first term.

$=\,\,$ $\dfrac{\sin^2{x}}{\cos{x}}$ $+$ $1 \times \dfrac{\cos{x}}{1}$

Writing the one as the quotient of cosine of angle $x$ divided by the cos of angle $x$ in the second term makes the denominator of the fraction in the second term will be the same as the denominator of the fraction in the first term.

$=\,\,$ $\dfrac{\sin^2{x}}{\cos{x}}$ $+$ $\dfrac{\cos{x}}{\cos{x}} \times \dfrac{\cos{x}}{1}$

The two fractions are multiplying in the second term and let’s find their product by the multiplication of the fractions.

$=\,\,$ $\dfrac{\sin^2{x}}{\cos{x}}$ $+$ $\dfrac{\cos{x} \times \cos{x}}{\cos{x} \times 1}$

Now, find the product of the factors in the both numerator and denominator of the second term.

$=\,\,$ $\dfrac{\sin^2{x}}{\cos{x}}$ $+$ $\dfrac{\cos^2{x}}{\cos{x}}$

Now, the two fractions have the same denominators and the two like fractions can be added by the addition of the like fractions.

$=\,\,$ $\dfrac{\sin^2{x}+\cos^2{x}}{\cos{x}}$

### Evaluate Trigonometric expression by simplifying

The sum of the squares of the sine and cosine is equal to one as per the pythagorean identity of sine and cosine functions.

$=\,\,$ $\dfrac{1}{\cos{x}}$

The reciprocal of cosine of angle $x$ is equal to secant of angle $x$, as per the reciprocal of cosine identity.

$=\,\,$ $\sec{x}$

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