# Evaluate :

## Question :

(i)  sin 60 cos 30 + cos 60 sin 30

(ii)  $$2 tan^2 45$$ + $$cos^2 30$$ – $$sin^2 60$$

(iii)  $$cos 45\over sec 30 + cosec 30$$

(iv)  $$sin 30 + tan 45 – cosec 60\over sec 30 + cos 60 + cot 45$$

(v)  $$5 cos^2 60 + 4 sec^2 30 – tan^2 45\over sin^2 30 + cos^2 30$$

## Solution :

(i)  sin 60 cos 30 + cos 60 sin 30

On substituting the values of various t-ratios in (1), we get

Given expression = $$\sqrt{3}\over 2$$ $$\times$$ $$\sqrt{3}\over 2$$ + $$1\over 2$$ $$\times$$ $$1\over 2$$ = $$4\over 4$$ = 1

(ii)  $$2 tan^2 45$$ + $$cos^2 30$$ – $$sin^2 60$$ = 2 $$\times$$ $$(1)^2$$ + $$({\sqrt{3}\over 2})^2$$ – $$({\sqrt{3}\over 2})^2$$ = 2

(iii)  $$cos 45\over sec 30 + cosec 30$$ = $${1\over \sqrt{2}}\over {2\over \sqrt{3}} + 2$$ = $$1\over \sqrt{2}$$ $$\times$$ $$\sqrt{3}\over 2 + 2\sqrt{3}$$

= $$\sqrt{3}\over \sqrt{2} \times 2(\sqrt{3} + 1)$$ $$\times$$ $$\sqrt{3} – 1\over \sqrt{3} – 1$$

= $$3 – \sqrt{3}\over 4\sqrt{2}$$.

(iv)  $$sin 30 + tan 45 – cosec 60\over sec 30 + cos 60 + cot 45$$ = $${1\over 2} + 1 – {2\over \sqrt{3}}\over {2\over \sqrt{3}} + {1\over 2} + 1$$

= $$3\sqrt{3} – 4\over 3\sqrt{3} + 4$$

(v)  $$5 cos^2 60 + 4 sec^2 30 – tan^2 45\over sin^2 30 + cos^2 30$$ = $${5\over 4} + {16\over 3} – 1\over {1\over 4} + {3\over 4}$$

= $${67\over 12}\over {4\over 4}$$ = $$67\over 12$$