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Question: Evaluate the following \[2{{\sin }^{2}}{{30}^{o}}-3{{\cos }^{2}}{{45}^{o}}+{{\tan }^{2}}{{60}^{o}}...

Evaluate the following
2sin⁡230o−3cos⁡245o+tan⁡260o2{{\sin }^{2}}{{30}^{o}}-3{{\cos }^{2}}{{45}^{o}}+{{\tan }^{2}}{{60}^{o}}

Explanation

Solution

Hint:First of all, consider the expression given in the question. Now make the table for trigonometric ratios of general angles. Now, from that find the values of sin⁡30o,cos⁡45o\sin {{30}^{o}},\cos {{45}^{o}} and tan⁡60o\tan {{60}^{o}} and substitute these in the given expression to get the required answer.

Complete step-by-step answer:
In this question, we have to find the value of the expression
E=2sin⁡230o−3cos⁡245o+tan⁡260o....(i)E=2{{\sin }^{2}}{{30}^{o}}-3{{\cos }^{2}}{{45}^{o}}+{{\tan }^{2}}{{60}^{o}}....\left( i \right).
Now, we have to find the values of sin⁡30o,cos⁡45o\sin {{30}^{o}},\cos {{45}^{o}} and tan⁡60o\tan {{60}^{o}}.
Let us make the table for trigonometric ratios of general angles like 0o,30o,45o,60o,90o{{0}^{o}},{{30}^{o}},{{45}^{o}},{{60}^{o}},{{90}^{o}} and find the required values.

From the above table, we get, sin⁡30o=12\sin {{30}^{o}}=\dfrac{1}{2}. By substituting this in equation (i), we get,
E=2(12)2−3cos⁡245o+tan⁡260oE=2{{\left( \dfrac{1}{2} \right)}^{2}}-3{{\cos }^{2}}{{45}^{o}}+{{\tan }^{2}}{{60}^{o}}
Also from the above table, we get cos⁡45o=12\cos {{45}^{o}}=\dfrac{1}{\sqrt{2}}. By substituting this in the above equation, we get, E=2(12)2−3(12)2+tan⁡260oE=2{{\left( \dfrac{1}{2} \right)}^{2}}-3{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\tan }^{2}}{{60}^{o}}
From the table, we also get, tan⁡60o=3\tan {{60}^{o}}=\sqrt{3}. By substituting this in the above equation, we get,
E=2(12)2−3(12)2+(3)2E=2{{\left( \dfrac{1}{2} \right)}^{2}}-3{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}}
By simplifying the above equation, we get,
E=2(14)−3(14)+3E=2\left( \dfrac{1}{4} \right)-3\left( \dfrac{1}{4} \right)+3
E=24−34+3E=\dfrac{2}{4}-\dfrac{3}{4}+3
E=12−34+3E=\dfrac{1}{2}-\dfrac{3}{4}+3
E=12−34+31E=\dfrac{1}{2}-\dfrac{3}{4}+\dfrac{3}{1}
E=2−3+124E=\dfrac{2-3+12}{4}
E=114E=\dfrac{11}{4}
Hence, we get the value of the expression 2sin⁡230o−3cos⁡245o+tan⁡260o2{{\sin }^{2}}{{30}^{o}}-3{{\cos }^{2}}{{45}^{o}}+{{\tan }^{2}}{{60}^{o}} as 114\dfrac{11}{4}.

Note: In these types of questions, first of all, it is very important for students to memorize the trigonometric table for general angles. Also, here students just need to remember the values of sin⁡θ\sin \theta and cos⁡θ\cos \theta at various angles like 30o,60o,45o,{{30}^{o}},{{60}^{o}},{{45}^{o}}, etc. and they can find tan⁡θ\tan \theta by using tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }. Also, students must take care of the calculation and solve the equation according to the BODMAS rule.