Question

Sin*30°-tan*-60°+cosec*60°+6sec*45°÷cosec90°+cos60°-tan*45°

Answer 1

Step-by-step explanation:

Given:

$sin^230$° $-tan^260$° $+cosec^260$° $+\frac{6sec^245^0}{cosec90^0}$$+cos60$° $-tan^245$°

To find:

The value of $sin^230$° $-tan^260$° $+cosec^260$° $+\frac{6sec^245^0}{cosec90^0}$ $+cos60$° $-tan^245$°

Solution:

Given that,

$sin^230$° $-tan^260$° $+cosec^290$° $+\frac{6sec^245^0}{cosec90^0}$ $+cos60$° $-tan^245$°

Using the trigonometric table values,

⇒ $(\frac{1}{\sqrt{2} })^2-(\sqrt{3})^2+(\frac{2}{\sqrt{3} })^2+ \frac{6(\sqrt{2})^2 }{1}+ \frac{1}{2} -(1)^2$

⇒ $\frac{1}{4}-3 +\frac{4}{3}+12 +\frac{1}{2} -1$

Add the whole numbers first,

⇒ $\frac{1}{4} +\frac{4}{3} +\frac{1}{2} +8$    

Find the LCD (Least Common Denominator) of $\frac{1}{4},\frac{4}{3},\frac{1}{2}$.

$LCD=12$

Make the denominators the same as the LCD.

⇒ $\frac{3}{12} +\frac{16}{12}+\frac{6}{12} +8$ × $\frac{12}{12}$

simplify the denominator are now the same,

⇒ $\frac{3}{12} +\frac{16}{12} +\frac{6}{12} +\frac{96}{12}$

Now join the denominators,

⇒ $\frac{3+16+6+96}{12}$

⇒ $\frac{121}{12}$

Convert to the mixed fraction and we get,

⇒ $10\frac{1}{2}$

Answer:

Hence the value of the given expression is $\frac{121}{12}$.