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Mathematics Question on Trigonometric Ratios of Some Specific Angles

Evaluate the following.
(i) sin60° cos30° + sin30° cos 60°
(ii) 2tan245° + cos230° - sin260°
(iii) cos45°sec30°+cosec30°\frac{cos 45°}{sec 30°+cosec30°}

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

(v) 5cos260°+4sec230°−tan245°sin230°+cos230°\frac{5cos^260°+4sec^230°-tan^245°}{sin^230°+cos^230°}

Answer

(i) sin60° cos30° + sin30° cos 60°

=32×32+(12)×(12)= \frac{\sqrt3}{2} ×\frac{\sqrt3}{2} + (\frac{1}{2}) ×(\frac{1}{2} )

=34+14=44=1=\frac{ 3}{4}+\frac{1}{4} =\frac{ 4}{4} =1

(ii) 2tan245° + cos230° - sin260°

=2(1)2\+(32)2−(32)2= 2(1)^2 \+ \left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt3}{2}\right)^2
= 2 + 0
= 2

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

=(12)[(23)+2]=\frac{\left (\frac{1}{\sqrt2}\right) }{\left [\left(\frac{2}{\sqrt3}\right) + 2\right]}

=(12)[(2+23)3]= \frac{\left(\frac{1}{\sqrt2}\right) }{ \left[\frac{(2 + 2\sqrt3)}{\sqrt3}\right]}

=(1×3)[2×(2+23)]=\frac{ (1 × \sqrt3) }{ \left[\sqrt2 × (2 + 2\sqrt3)\right]}

=3[22(3+1)]= \frac{\sqrt3 }{ \left[2\sqrt2(\sqrt3 + 1)\right]}

Multiplying numerator and denominator by 2(3−1)\sqrt2 (\sqrt3 - 1), we get

=3[22(3+1)]×2(3−1)2(3−1)= \frac{\sqrt3 }{ \left[2\sqrt2(\sqrt3 + 1)\right]} × \frac{\sqrt2 (\sqrt3 - 1) }{ \sqrt2 (\sqrt3 - 1)}

=(32−6)4(3−1)=\frac{ (3\sqrt2 - \sqrt6) }{ 4(3 - 1)}

=(32−6)8=\frac{ (3\sqrt2 - \sqrt6) }{ 8}

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

=[12+1−2323+12+1]= \left[\frac{\frac{1}{2} + 1 - \frac{2}{\sqrt3} }{ \frac{2}{\sqrt3} + \frac{1}{2} + 1}\right]

=32−2323+32=\frac{ \frac{3}{2} - \frac{2}{\sqrt3} }{ \frac{2}{\sqrt3} + \frac{3}{2}}

=[33−4234+3323]= \left[\frac{{\frac{3\sqrt3 - 4}{2\sqrt3}} }{ {\frac{4 + 3\sqrt3}{2\sqrt3}}}\right]

=(33−4)(33+4)=\frac{ (3\sqrt3 - 4) }{ (3\sqrt3 + 4)}

Multiplying numerator and denominator by 33−43\sqrt3 - 4, we get

=(33−4)(33−4)(33+4)(33−4)=\frac{ (3\sqrt3 - 4)(3\sqrt3 - 4) }{ (3\sqrt3 + 4)(3\sqrt3 - 4)}

=(27+16−243)(27−16)= \frac{(27 + 16 - 24\sqrt3) }{ (27 - 16)}

=(43−243)11= \frac{(43 - 24\sqrt3) }{ 11}

**(v) **(5cos260°+4sec230°−tan245°)(sin230°+cos230°)\frac{ (5cos^2 60° + 4sec^2 30° - tan^2 45°) }{ (sin^2 30° + cos^2 30°)}

=[5×(12)2\+4×(23)2\-(1)2][(12)2\+(32)2]= \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]}

=(54+163−1)(14+34)= \frac{(\frac{5}{4} + \frac{16}{3} - 1) }{ (\frac{1}{4} +\frac{ 3}{4})}

=[15+64−1212][3+14]=\frac{\left [\frac{15 + 64 - 12}{12}\right] }{ \left[\frac{3 + 1}{4}\right]}

=671244= \frac{\frac{67}{12} }{ \frac{4}{4}}

=6712= \frac{67}{12}