Author name: mathemerize

Solution : The value of sin 72 degrees is \(\sqrt{10 + 2\sqrt{5}}\over 4\). Proof : We know that value of cos 18 degrees is \(\sqrt{10 + 2\sqrt{5}}\over 4\). Since 72 degree is the complement of 18 degree. \(\therefore\) sin 72 = sin(90 – 18) = cos 18 = \(\sqrt{10 + 2\sqrt{5}}\over 4\) Hence, sin 72 […]

Solution : The value of cos 18 degrees is \(\sqrt{10 + 2\sqrt{5}}\over 4\). Proof : We know that the value of sin 18 degrees is \(\sqrt{5} – 1\over 4\). Putting \(\theta\) = 18 in  \(cos \theta\) = \(\sqrt{1 – sin^2 \theta}\), we get cos 18 = \(\sqrt{1 – sin^2 18}\) = \(\sqrt{1 – ({\sqrt{5} –

Solution : The value of sin 18 degrees is \(\sqrt{5} – 1\over 4\). Proof :  Let \(\theta\) = 18 degrees. Then, \(5\theta\) = 90 degrees \(\implies\)   \(2\theta\) + \(3\theta\) = 90 \(\implies\)   \(2\theta\) = 90 – \(3\theta\) \(\implies\)   \(sin 2\theta\) = \(sin (90 – 3\theta)\) \(\implies\)  \(sin 2\theta\) = \(cos 3\theta\) By using the formula

Solution : We have, L.H.S = cot A + cot (60 + A) –  cot (60 – A) \(\implies\) L.H.S = \(1\over tan A\) + \(1\over tan (60 + A)\) – \(1\over tan (60 – A)\) \(\implies\) L.H.S = \(1\over tan A\)+ \(1 – \sqrt{3} tan A\over \sqrt{3} + tan A\) – \(1 + \sqrt{3}

Solution : We have, L.H.S = tan A + tan (60 + A) –  tan (60 – A) \(\implies\) L.H.S = tan A + \(\sqrt{3} + tan A\over 1 – \sqrt{3} tan A\) – \(\sqrt{3} – tan A\over 1 + \sqrt{3} tan A\) [ By using this formula, tan (A + B) = \(tan A

Solution : We have, L.H.S = cos A cos (60 – A) cos (60 + A) \(\implies\) L.H.S = cos A (\(cos^2 60 – sin^2 A\)) [ By using this formula, cos (A + B) cos (A – B) = \(cos^2 A\) – \(sin^2 B\)  above ] \(\implies\) L.H.S = cos A (\(1\over 4\) –

Solution : We have, L.H.S = sin A sin (60 – A) sin (60 + A) \(\implies\) L.H.S = sin A (\(sin^2 60 – sin^2 A\)) [ By using this formula, sin (A + B) sin (A – B) = \(sin^2 A\) – \(sin^2 B\)  above ] \(\implies\) L.H.S = sin A (\(3\over 4\) –