Find the value of the following without using tables:
$\tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }$
Answer
617.1k+ views
Hint: First convert $\tan {60^\circ }$ in terms of $\cot$ and then write $\cot {60^\circ }$ in terms of $\tan {30^\circ }$ to easily evaluate the first terms. Now write the value of $\cos {0^\circ }$ as it is and then convert $\sec$ into $\cos$ 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,
$\Rightarrow \tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }$
Firstly, convert $\tan {60^\circ }$ in terms of $\cot$.
We can do this by using the conversion, $\tan \theta = \dfrac{1}{{\cot \theta }}$
$\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\cot {{60}^\circ }}} + \cos {0^\circ } + \sec {60^\circ }$
We shall now convert $\cot {60^\circ }$ to $\tan {30^\circ }$ to easily evaluate.
We can do this by using, $\cot \theta = \tan (90 - \theta )$
$\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\tan ({{90}^\circ } - {{60}^\circ })}} + \cos {0^\circ } + \sec {60^\circ }$
On further evaluation,
$\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\tan ({{30}^\circ })}} + \cos {0^\circ } + \sec {60^\circ }$
Simplify the first term to get,
$\Rightarrow 1 + \cos {0^\circ } + \sec {60^\circ }$
Write the value of $\cos {0^\circ }$ as it is since there is no further simplification in that term.
$\Rightarrow 1 + 1 + \sec {60^\circ }$
Now convert $\sec$ into $\cos$
$\Rightarrow 1 + 1 + \dfrac{1}{{\cos {{60}^\circ }}}$
Write the value of $\cos {60^\circ }$ in the expression and evaluate.
$\Rightarrow 1 + 1 + \dfrac{1}{{\dfrac{1}{2}}}$
$\Rightarrow 1 + 1 + 2$
On further evaluation we get,
$\Rightarrow 4$
$\therefore$ $\tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }$ is equal to the value $4$.
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 $\cos \theta$ or $\sin \theta$ . 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 ${1^\circ } \times \dfrac{\pi }{{180}} = 0.017Rad$ . Express everything in $\sin \theta$ or $\sin \theta$ 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.
Complete step by step solution:
The given expression is,
$\Rightarrow \tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }$
Firstly, convert $\tan {60^\circ }$ in terms of $\cot$.
We can do this by using the conversion, $\tan \theta = \dfrac{1}{{\cot \theta }}$
$\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\cot {{60}^\circ }}} + \cos {0^\circ } + \sec {60^\circ }$
We shall now convert $\cot {60^\circ }$ to $\tan {30^\circ }$ to easily evaluate.
We can do this by using, $\cot \theta = \tan (90 - \theta )$
$\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\tan ({{90}^\circ } - {{60}^\circ })}} + \cos {0^\circ } + \sec {60^\circ }$
On further evaluation,
$\Rightarrow \tan {30^\circ } \times \dfrac{1}{{\tan ({{30}^\circ })}} + \cos {0^\circ } + \sec {60^\circ }$
Simplify the first term to get,
$\Rightarrow 1 + \cos {0^\circ } + \sec {60^\circ }$
Write the value of $\cos {0^\circ }$ as it is since there is no further simplification in that term.
$\Rightarrow 1 + 1 + \sec {60^\circ }$
Now convert $\sec$ into $\cos$
$\Rightarrow 1 + 1 + \dfrac{1}{{\cos {{60}^\circ }}}$
Write the value of $\cos {60^\circ }$ in the expression and evaluate.
$\Rightarrow 1 + 1 + \dfrac{1}{{\dfrac{1}{2}}}$
$\Rightarrow 1 + 1 + 2$
On further evaluation we get,
$\Rightarrow 4$
$\therefore$ $\tan {30^\circ } \times \tan {60^\circ } + \cos {0^\circ } + \sec {60^\circ }$ is equal to the value $4$.
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 $\cos \theta$ or $\sin \theta$ . 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 ${1^\circ } \times \dfrac{\pi }{{180}} = 0.017Rad$ . Express everything in $\sin \theta$ or $\sin \theta$ 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.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

