Question

Evaluate the following. 
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan²45° + cos²30° - sin²60°
(iii) \(\frac{cos 45°}{sec 30°+cosec30°}\)

(iv) \(\frac{sin\ 30°+tan\ 45°cosec\ 60°}{sec\ 30°+cos\ 60°+cot\ 45°}\)

(v) \(\frac{5cos^260°+4sec^230°-tan^245°}{sin^230°+cos^230°}\)

Answer 1

(i) sin60° cos30° + sin30° cos 60°

\(= \frac{\sqrt3}{2} ×\frac{\sqrt3}{2} + (\frac{1}{2}) ×(\frac{1}{2} )\)

\(=\frac{ 3}{4}+\frac{1}{4} =\frac{ 4}{4} =1\)

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(ii) 2tan²45° + cos²30° - sin²60°

\(= 2(1)^2 + \left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt3}{2}\right)^2\)
 = 2 + 0
= 2

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(iii) \(\frac{cos\ 45°}{(sec\ 30°+cosec\ 30°)}\)

\(=\frac{\left (\frac{1}{\sqrt2}\right) }{\left [\left(\frac{2}{\sqrt3}\right) + 2\right]}\)

\(= \frac{\left(\frac{1}{\sqrt2}\right) }{ \left[\frac{(2 + 2\sqrt3)}{\sqrt3}\right]}\)

\(=\frac{ (1 × \sqrt3) }{ \left[\sqrt2 × (2 + 2\sqrt3)\right]}\)

\(= \frac{\sqrt3 }{ \left[2\sqrt2(\sqrt3 + 1)\right]}\)

Multiplying numerator and denominator by \(\sqrt2 (\sqrt3 - 1)\), we get

\(= \frac{\sqrt3 }{ \left[2\sqrt2(\sqrt3 + 1)\right]} × \frac{\sqrt2 (\sqrt3 - 1) }{ \sqrt2 (\sqrt3 - 1)}\)

\(=\frac{ (3\sqrt2 - \sqrt6) }{ 4(3 - 1)}\)

\(=\frac{ (3\sqrt2 - \sqrt6) }{ 8}\)

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(iv) \(\frac{(sin\  30° + tan\  45° - cosec\ 60°) }{(sec\  30° + cos\ 60° + cot\ 45°)}\)

\(= \left[\frac{\frac{1}{2} + 1 - \frac{2}{\sqrt3} }{ \frac{2}{\sqrt3} + \frac{1}{2} + 1}\right]\)

\(=\frac{ \frac{3}{2} - \frac{2}{\sqrt3} }{ \frac{2}{\sqrt3} + \frac{3}{2}}\)

\(= \left[\frac{{\frac{3\sqrt3 - 4}{2\sqrt3}} }{ {\frac{4 + 3\sqrt3}{2\sqrt3}}}\right]\)

\(=\frac{ (3\sqrt3 - 4) }{ (3\sqrt3 + 4)}\)

Multiplying numerator and denominator by \(3\sqrt3 - 4\), we get

\(=\frac{ (3\sqrt3 - 4)(3\sqrt3 - 4) }{ (3\sqrt3 + 4)(3\sqrt3 - 4)}\)

\(= \frac{(27 + 16 - 24\sqrt3) }{ (27 - 16)}\)

\(= \frac{(43 - 24\sqrt3) }{ 11}\)

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(v) \(\frac{ (5cos^2 60° + 4sec^2 30° - tan^2 45°) }{ (sin^2 30° + cos^2 30°)}\)

\(= \frac{\left[5 × \left(\frac{1}{2}\right)^2 + 4 × \left(\frac{2}{\sqrt3}\right)^2 - (1)^2\right] }{\left [\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt3}{2}\right)^2\right]}\)

\(= \frac{(\frac{5}{4} + \frac{16}{3} - 1) }{ (\frac{1}{4} +\frac{ 3}{4})}\)

\(=\frac{\left [\frac{15 + 64 - 12}{12}\right] }{ \left[\frac{3 + 1}{4}\right]}\)

\(= \frac{\frac{67}{12} }{ \frac{4}{4}}\)

\(= \frac{67}{12}\)