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Question: Evaluate each of the following: i) \(\sin {30^ \circ } + \cos {45^ \circ } + \tan {180^ \circ }\) ...

Evaluate each of the following:
i) sin⁡30∘+cos⁡45∘+tan⁡180∘\sin {30^ \circ } + \cos {45^ \circ } + \tan {180^ \circ }
ii) cosec⁡45∘+cot⁡45∘+tan⁡0∘\operatorname{cosec} {45^ \circ } + \cot {45^ \circ } + \tan {0^ \circ }
iii) sin⁡30∘×cos⁡45∘×tan⁡360∘\sin {30^ \circ } \times \cos {45^ \circ } \times \tan {360^ \circ }

Explanation

Solution

Write the values of the given trigonometric ratios, like, sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2}, cos⁡45∘=12\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }} . then, substitute these values in the given expressions. Simplify the equations to find the value of a given expression.

Complete step-by-step answer:
In part (i), we have to find the value of sin⁡30∘+cos⁡45∘+tan⁡180∘\sin {30^ \circ } + \cos {45^ \circ } + \tan {180^ \circ }
We know that sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2}, cos⁡45∘=12\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }} and tan⁡180∘=tan⁡(180∘−0)=−tan⁡0=0\tan {180^ \circ } = \tan \left( {{{180}^ \circ } - 0} \right) = - \tan 0 = 0
On substituting the values in part (i), we will get,
12+12+0=2+222=2(1+2)22=1+22\dfrac{1}{2} + \dfrac{1}{{\sqrt 2 }} + 0 = \dfrac{{\sqrt 2 + 2}}{{2\sqrt 2 }} = \dfrac{{\sqrt 2 \left( {1 + \sqrt 2 } \right)}}{{2\sqrt 2 }} = \dfrac{{1 + \sqrt 2 }}{2}
In part (ii), we have to find the value of cosec⁡45∘+cot⁡45∘+tan⁡0∘\operatorname{cosec} {45^ \circ } + \cot {45^ \circ } + \tan {0^ \circ }
We know that cosec⁡45∘=2\operatorname{cosec} {45^ \circ } = \sqrt 2 , cot⁡45∘=1\cot {45^ \circ } = 1 and tan⁡0∘=0\tan {0^ \circ } = 0
We will substitute the values in the equation, cosec⁡45∘+cot⁡45∘+tan⁡0∘\operatorname{cosec} {45^ \circ } + \cot {45^ \circ } + \tan {0^ \circ }
2+1+0=2+1\sqrt 2 + 1 + 0 = \sqrt 2 + 1
In part (iii), we will find the value of sin⁡30∘×cos⁡45∘×tan⁡360∘\sin {30^ \circ } \times \cos {45^ \circ } \times \tan {360^ \circ }
Again, we have sin⁡30∘=12\sin {30^ \circ } = \dfrac{1}{2}, cos⁡45∘=12\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }} and tan⁡360∘=tan⁡(360∘−0)=−tan⁡0=0\tan {360^ \circ } = \tan \left( {{{360}^ \circ } - 0} \right) = - \tan 0 = 0
On substituting the values in part (iii), we will get,
12×12×0=0\dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} \times 0 = 0

Note: For these types of questions, a student must know the values of trigonometric ratios of certain angles. The value of the angles that lies in the first quadrant is positive for all ratios, in second quadrant sine and cosecant value is positive, in third quadrant tangent and cotangent value is positive, and in fourth quadrant, value of cosine and secant is positive.