Question

Evaluate exactly without the use of a calculator (remember to rationalize any denominators). a.(cos 60o)(sin 270o) + tan 225o b. – tan 240o + (cos 45o)(sec 135o)

Answer 1

Given:

The trigonometric expressions are given as,

$\begin{gathered} a)\text{ }(\cos 60\degree)(\sin 270\degree)+\tan 225\degree \\ b)-\tan 240\degree+(cos45\degree)(\sec 135\degree) \end{gathered}$

Explanation:

a)

The given expression can be rewritten as,

$=(\cos 60\degree)(\sin (360\degree-90\degree))+\tan (270\degree-45\degree)\text{ . . . . .(1)}$

Since, from the trigonometric ratios,

$\begin{gathered} \sin (360\degree-90\degree)=-\sin 90\degree \\ \tan (270\degree-45\degree)=\cot 45\degree \end{gathered}$

On plugging the obtained ratios in equation (1),

$=(\cos 60\degree)(-\sin 90\degree)+\cot 45$

Substitute the trigonometric values in the above equation.

$\begin{gathered} =\frac{1}{2}(-1)+1 \\ =-\frac{1}{2}+1 \\ =\frac{1}{2} \end{gathered}$

Hence, the exact value of the expression is 1/2.

b)

The given expression can be rewritten as,

$=-\tan (270\degree-30\degree)+(\cos 45\degree)(\sec (90\degree+45\degree))\text{ . . . ..(2)}$

Since, from the trigonometric ratios,

$\begin{gathered} \tan (270\degree-30\degree)=\cot 30\degree \\ \sec (90\degree+45\degree)=-\csc 45\degree \end{gathered}$

On plugging the obtained ratios in equation (2),

$=-\cot 30\degree+(\cos 45\degree)(-\csc 45)$

Substitute the trigonometric values in the above equation.

$\begin{gathered} =-\sqrt[]{3}+\frac{1}{\sqrt[]{2}}(-\sqrt[]{2}) \\ =-\sqrt[]{3}-1 \end{gathered}$

Hence, the exact value of the expression is -√3-1.