Without using the trigonometric tables find the value of the following expressions:
$\dfrac{{\sec ({{90}^0}{\text{ - }}\theta {\text{)cosec}}\theta {\text{ - tan(9}}{{\text{0}}^0}{\text{ - }}\theta )\cot \theta {\text{ + co}}{{\text{s}}^2}{{25}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\tan {{63}^0}}}$

Hint: To solve this question we will use the trigonometric properties like $\tan \theta {\text{ = }}\dfrac{1}{{\cot \theta }}$, $\sec \theta {\text{ = }}\dfrac{1}{{\cos \theta }}$ and properties related to angles like $\cos ({90^0} - {\text{ }}\theta {\text{) = sin}}\theta $. Now, we will use some trigonometric properties to solve the given question without using the trigonometric table.
From trigonometry, we know that $\cos ({90^0} - {\text{ }}\theta {\text{) = sin}}\theta $, $\sec ({90^0} - {\text{ }}\theta {\text{) = cosec}}\theta $,$\tan ({90^0} - {\text{ }}\theta {\text{) = cot}}\theta $. Also, $\tan \theta {\text{ = }}\dfrac{1}{{\cot \theta }}$. We will use all these properties to solve the given question

Complete step-by-step solution -

Now, we are given $\dfrac{{\sec ({{90}^0}{\text{ - }}\theta {\text{)cosec}}\theta {\text{ - tan(9}}{{\text{0}}^0}{\text{ - }}\theta )\cot \theta {\text{ + co}}{{\text{s}}^2}{{25}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\tan {{63}^0}}}$. So, it can be written as,
$\dfrac{{\cos ec\theta {\text{cosec}}\theta {\text{ - cot}}\theta \cot \theta {\text{ + co}}{{\text{s}}^2}{{25}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\tan {{63}^0}}}$
$ \Rightarrow $ $\dfrac{{\cos e{c^2}\theta {\text{ - co}}{{\text{t}}^2}\theta {\text{ + co}}{{\text{s}}^2}({{90}^0}{\text{ - 6}}{{\text{5}}^0}{\text{) + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\tan ({{90}^0}{\text{ - 2}}{{\text{7}}^0})}}$
$ \Rightarrow $ $\dfrac{{\cos e{c^2}\theta {\text{ - co}}{{\text{t}}^2}\theta {\text{ + si}}{{\text{n}}^2}{\text{6}}{{\text{5}}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\cot {\text{2}}{{\text{7}}^0}}}$
Now, we know that ${\sin ^2}\theta {\text{ + co}}{{\text{s}}^2}\theta {\text{ = 1}}$ and $\cos e{c^2}\theta {\text{ - co}}{{\text{t}}^2}\theta {\text{ = 1}}$. Therefore, applying these identities in the above equation, we get
$\dfrac{{\cos e{c^2}\theta {\text{ - co}}{{\text{t}}^2}\theta {\text{ + si}}{{\text{n}}^2}{\text{6}}{{\text{5}}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\cot {\text{2}}{{\text{7}}^0}}}$ = $\dfrac{{1{\text{ + }}{\text{1}}}}{{3\tan {{27}^0}.\cot {{27}^0}}}$
As, $\tan \theta {\text{ = }}\dfrac{1}{{\cot \theta }}$. So,
$\dfrac{{\cos e{c^2}\theta {\text{ - co}}{{\text{t}}^2}\theta {\text{ + si}}{{\text{n}}^2}{\text{6}}{{\text{5}}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\cot {\text{2}}{{\text{7}}^0}}}$ = $\dfrac{{1{\text{ + 1}}}}{3}{\text{ = }}\dfrac{2}{3}$
Therefore, $\dfrac{{\sec ({{90}^0}{\text{ - }}\theta {\text{)cosec}}\theta {\text{ - tan(9}}{{\text{0}}^0}{\text{ - }}\theta )\cot \theta {\text{ + co}}{{\text{s}}^2}{{25}^0}{\text{ + co}}{{\text{s}}^2}{{65}^0}}}{{3\tan {{27}^0}.\tan {{63}^0}}}$ = $\dfrac{2}{3}$

Note: When we come up with such types of questions, we have to use trigonometric identities and properties to solve the question. In the above question we have use the identity $\cos ({90^0} - {\text{ }}\theta {\text{) = sin}}\theta $ which is only applicable when the angle $\theta $ is less than ${90^0}$ and lies in the first quadrant. If the angle is greater than ${90^0}$, then we have to use a different formula, which can be derived from the formula cos (x + y) = cosxcosy – sinxsiny. Keeping angle in place of y and the reference angle (${90^0}$in the first quadrant) in place of x, any formula for any quadrant can be found.