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Question: Find the value of the following without using tables: \(\tan {30^\circ } \times \tan {60^\circ } +...

Find the value of the following without using tables:
tan30×tan60+cos0+sec60\tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }

Explanation

Solution

First convert tan60\tan {60^\circ } in terms of cot\cot and then write cot60\cot {60^\circ } in terms of tan30\tan {30^\circ } to easily evaluate the first terms. Now write the value of cos0\cos {0^\circ } as it is and then convert sec\sec into cos\cos and then find the value of it. Here we must use tables for the last two terms because without which we cannot find the value of the expression.

Complete step by step solution:
The given expression is,
tan30×tan60+cos0+sec60\Rightarrow \tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }
Firstly, convert tan60\tan {60^\circ } in terms of cot\cot.
We can do this by using the conversion, tanθ=1cotθ\tan \theta = \dfrac{1}{{\cot \theta }}
tan30×1cot60+cos0+sec60\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\cot {{60}^\circ }}} + \cos {0^\circ } + \sec {60^\circ }
We shall now convert cot60\cot {60^\circ } to tan30\tan {30^\circ } to easily evaluate.
We can do this by using, cotθ=tan(90θ)\cot \theta = \tan (90 - \theta )
tan30×1tan(9060)+cos0+sec60\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\tan ({{90}^\circ } - {{60}^\circ })}} + \cos {0^\circ } + \sec {60^\circ }
On further evaluation,
tan30×1tan(30)+cos0+sec60\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\tan ({{30}^\circ })}} + \cos {0^\circ } + \sec {60^\circ }
Simplify the first term to get,
1+cos0+sec60\Rightarrow 1 + \cos {0^\circ } + \sec {60^\circ }
Write the value of cos0\cos {0^\circ } as it is since there is no further simplification in that term.
1+1+sec60\Rightarrow 1 + 1 + \sec {60^\circ }
Now convert sec\sec into cos\cos
1+1+1cos60\Rightarrow 1 + 1 + \dfrac{1}{{\cos {{60}^\circ }}}
Write the value of cos60\cos {60^\circ } in the expression and evaluate.
1+1+112\Rightarrow 1 + 1 + \dfrac{1}{{\dfrac{1}{2}}}
1+1+2\Rightarrow 1 + 1 + 2
On further evaluation we get,
4\Rightarrow 4

\therefore tan30×tan60+cos0+sec60\tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ } is equal to the value 44.

Additional Information: Whenever complex equations are given to solve one must always Firstly start from the complex side and then convert all the terms into cosθ\cos \theta or sinθ\sin \theta . Then combine them into single fractions. Now it’s most likely to use Trigonometric identities for the transformations if there are any. Know when and where to apply the Subtraction-Addition formula.

Note: Always check when the trigonometric functions are given in degrees or radians. There’s a lot of difference between both 1×π180=0.017Rad{1^\circ } \times \dfrac{\pi }{{180}} = 0.017Rad . Express everything in sinθ\sin \theta or sinθ\sin \theta to easily evaluate. It is a must to memorize the values of basic trigonometric functions since all the functions can be written in terms of those basic trigonometric functions and can be easily evaluated.