Question

Evaluate $\dfrac{{{\tan }^{2}}{{60}^{0}}+4{{\cos }^{2}}{{45}^{0}}+3{{\sec }^{2}}{{30}^{0}}+5{{\cos }^{2}}{{90}^{0}}}{\csc {{30}^{0}}+\sec {{60}^{0}}-{{\cot }^{2}}{{30}^{0}}}$

Answer 1

Hint: The trigonometric ratios table helps to find the values of trigonometric standard angles such as ${{0}^{0}},{{30}^{0}},{{45}^{0}},{{60}^{0}}$ and ${{90}^{0}}$. It consists of trigonometric ratios – sine, cosine, tangent, cosecant, secant and cotangent. These ratios can be written in short as sin, cos, tan, csc, sec and cot.

Complete step-by-step answer:
The calculus is based on trigonometry and algebra. The fundamental trigonometric functions like sine and cosine are used to describe the sound and light waves. Trigonometry is used in oceanography to calculate heights of waves and tides in oceans. It is used in satellite systems.

The value of the trigonometric ratios by using the trigonometric table is given below.
$\tan {{60}^{0}}=\sqrt{3},\cos {{45}^{0}}=\dfrac{1}{\sqrt{2}},\sec {{30}^{0}}=\dfrac{2}{\sqrt{3}}\cos {{90}^{0}}=0,\csc {{30}^{0}}=2,\sec {{60}^{0}}=2,\cot 30=\sqrt{3}$

Let us consider the given expression and put all the trigonometric values, we get
$\dfrac{{{\tan }^{2}}{{60}^{0}}+4{{\cos }^{2}}{{45}^{0}}+3{{\sec }^{2}}{{30}^{0}}+5{{\cos }^{2}}{{90}^{0}}}{\csc {{30}^{0}}+\sec {{60}^{0}}-{{\cot }^{2}}{{30}^{0}}}=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+4{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+3{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}+5{{\left( 0 \right)}^{2}}}{2+2-{{\left( \sqrt{3} \right)}^{2}}}$

Simplifying the terms on right side, we get
$\dfrac{{{\tan }^{2}}{{60}^{0}}+4{{\cos }^{2}}{{45}^{0}}+3{{\sec }^{2}}{{30}^{0}}+5{{\cos }^{2}}{{90}^{0}}}{\csc {{30}^{0}}+\sec {{60}^{0}}-{{\cot }^{2}}{{30}^{0}}}=\dfrac{3+4\times \dfrac{1}{2}+3\times \dfrac{4}{3}+0}{2+2-3}$

Rearranging the terms on right side, we get
$\dfrac{{{\tan }^{2}}{{60}^{0}}+4{{\cos }^{2}}{{45}^{0}}+3{{\sec }^{2}}{{30}^{0}}+5{{\cos }^{2}}{{90}^{0}}}{\csc {{30}^{0}}+\sec {{60}^{0}}-{{\cot }^{2}}{{30}^{0}}}=\dfrac{3+2+4}{4-3}$

Again, rearranging the terms on right side, we get
$\dfrac{{{\tan }^{2}}{{60}^{0}}+4{{\cos }^{2}}{{45}^{0}}+3{{\sec }^{2}}{{30}^{0}}+5{{\cos }^{2}}{{90}^{0}}}{\csc {{30}^{0}}+\sec {{60}^{0}}-{{\cot }^{2}}{{30}^{0}}}=\dfrac{9}{1}$
$\dfrac{{{\tan }^{2}}{{60}^{0}}+4{{\cos }^{2}}{{45}^{0}}+3{{\sec }^{2}}{{30}^{0}}+5{{\cos }^{2}}{{90}^{0}}}{\csc {{30}^{0}}+\sec {{60}^{0}}-{{\cot }^{2}}{{30}^{0}}}=9$
Hence the required value of the given expression is 9.

Note: SOHCAHTOA, A way of remembering how to compute the sine, cosine, and tangent of an angle. SOH stands for Sine equals Opposite over Hypotenuse. CAH stands for Cosine equals Adjacent over Hypotenuse. TOA stands for Tangent equals Opposite over Adjacent.