Question

Show that:

(i) `2(cos^2 45º + tan^2 60º) – 6(sin^2 45º – tan^2 30º) = 6`

(ii) `2(cos^4 60º + sin^4 30º) – (tan^2 60º + cot^2 45º) + 3 sec^2 30º = 1/4`

Answer 1

(i) `2(cos^2 45º + tan^2 60º) – 6(sin^2 45º – tan^2 30º)`

`=2( ( \frac{1}{\sqrt{2}})^{2}+(\sqrt{3})^{2})-6(( \frac{1}{\sqrt{2}})^{2}-(\frac{1}{\sqrt{3}})^{2})`

`=2( \frac{1}{2}+3)-6( \frac{1}{2}-\frac{1}{3})=2( \frac{1+6}{2})-6( \frac{3-2}{6})`

`=2\times \frac{7}{2}-6\times \frac{1}{6}=7-1=6`

(ii)

` 2(cos^4 60º + sin^4 30º) – (tan^2 60º + cot^2 45º) + 3 sec^2 30º`

`=2( ( \frac{1}{2})^{4}+( \frac{1}{2})^{4})-( (\sqrt{3})^{2}+(1)^{2})+3( \frac{2}{\sqrt{3}})^{2}`

`=2( \frac{1}{16}+\frac{1}{16})( 3+1 )+3\times\frac{4}{3`

`=2\times \frac{1}{8}-4+4=\frac{1}{4}`