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) \(5 cos^2 60 + 4 sec^2 30 – tan^2 45\over sin^2 30 + cos^2 30\) = \({5\over 4} + {16\over 3} – 1\over {1\over 4} + {3\over 4}\)
= \({67\over 12}\over {4\over 4}\) = \(67\over 12\)
