Find the answer
\[{\sec ^2}10 - {\cot ^2}80 + \dfrac{{\sin 15\cos 75 + \cos 15\sin 75}}{{\cos \theta \sin \left( {90 - \theta } \right) + \sin \theta \cos \left( {90 - \theta } \right)}}\]

Hint:
We will simplify the expression by using the formula for the sine of the sum of 2 angles. We will further simplify it using various properties of trigonometric functions. We will solve the expression after we obtain integral results and calculate the answer.

Formulas used:
1) Sine of the sum of 2 angles (say \[A\] and \[B\]) is given by the sum of the product of Sine of \[A\] and Cosine of \[B\] and the product of Cosine of \[A\] and Sine of \[B\]:
\[ \Rightarrow {\mathop{\rm Sin}\nolimits} \left( {A + B} \right) = {\mathop{\rm Sin}\nolimits} A{\mathop{\rm Cos}\nolimits} B + {\mathop{\rm Sin}\nolimits} B{\mathop{\rm Cos}\nolimits} A\]
2) Secant of an angle is the reciprocal of Cosine of that angle:
\[ \Rightarrow \sec x = \dfrac{1}{{\cos x}}\]
3) Cotangent of an angle is the ratio of the cosine of that angle and the sine of that angle:
\[ \Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}}\]
4) Sine of the difference of an angle (say \[x\]) from 90 is the same as the cosine of that an angle:
\[ \Rightarrow \sin \left( {90 - x} \right) = \cos x\]
5) Cosine of the difference of an angle (say \[x\]) from 90 is the same as the sine of that an angle:
\[ \Rightarrow \cos \left( {90 - x} \right) = \sin x\]
6) Sum of the square of the sine of an angle and the cosine of the same angle is 1:
\[\begin{array}{l} \Rightarrow {\sin ^2}x + {\cos ^2}x = 1\\ \Rightarrow {\rm{ }}{\sin ^2}x = 1 - {\cos ^2}x\end{array}\]

Complete step by step solution:
We will evaluate the right-most term first. We will substitute 15 for \[A\] and 75 for \[B\] in the formula for the Sine of the sum of 2 angles in the numerator:
\[ \Rightarrow {\sec ^2}10 - {\cot ^2}80 + \dfrac{{\sin \left( {15 + 75} \right)}}{{\cos \theta \sin \left( {90 - \theta } \right) + \sin \theta \cos \left( {90 - \theta } \right)}}\]
We will substitute \[\theta \] for \[A\]and \[\left( {90 - \theta } \right)\] for \[B\] in the formula for the Sine of the sum of 2 angles in the denominator:
\[\begin{array}{l} \Rightarrow {\sec ^2}10 - {\cot ^2}80 + \dfrac{{\sin \left( {15 + 75} \right)}}{{\sin \left( {\theta + 90 - \theta } \right)}}\\ \Rightarrow {\sec ^2}10 - {\cot ^2}80 + \dfrac{{\sin 90}}{{\sin 90}}\\ \Rightarrow {\sec ^2}10 - {\cot ^2}80 + 1\end{array}\]
We will substitute 10 for \[x\] in the formula for the secant of an angle:
\[ \Rightarrow \dfrac{1}{{{{\cos }^2}10}} - {\cot ^2}80 + 1\]
We will substitute 80 for \[x\] in the formula for the cotangent of an angle:
\[\begin{array}{l} \Rightarrow \dfrac{1}{{{{\cos }^2}10}} - \dfrac{{{{\cos }^2}80}}{{{{\sin }^2}80}} + 1\\ \Rightarrow \dfrac{1}{{{{\cos }^2}\left( {90 - 80} \right)}} - \dfrac{{{{\cos }^2}80}}{{{{\sin }^2}\left( {90 - 10} \right)}} + 1\end{array}\]
We will substitute 10 for \[x\] in the formula for the Sine of difference of an angle from 90:
\[\begin{array}{l} \Rightarrow \dfrac{1}{{{{\cos }^2}\left( {90 - 80} \right)}} - \dfrac{{{{\cos }^2}80}}{{{{\sin }^2}80}} + 1\\ \Rightarrow \dfrac{1}{{{{\cos }^2}\left( {90 - 80} \right)}} - \dfrac{{{{\cos }^2}80}}{{{{\cos }^2}10}} + 1\end{array}\]
We will substitute 80 for \[x\] in the formula for the sum of square of the sine and the cosine of an angle:
\[ \Rightarrow \dfrac{{{{\sin }^2}80}}{{{{\cos }^2}\left( {90 - 80} \right)}} + 1\]
We will substitute 80 for \[x\] in the formula for the Cosine of the difference of an angle from 90:
\[\begin{array}{l} \Rightarrow \dfrac{{{{\sin }^2}80}}{{{{\sin }^2}80}} + 1\\ \Rightarrow 1 + 1\\ \Rightarrow 2\end{array}\]

\[\therefore \] The answer is 2.

Note:
We must know that the trigonometric functions are defined as ratios of different sides of a right-angled triangle. Sine is the ratio of the perpendicular and hypotenuse, cosine is the ratio of the base and hypotenuse, tangent is the ratio of perpendicular and base and so on.