JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 15)
If $${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5},$$ then
$${\tan ^2}x = {2 \over 3}$$
$${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {1 \over {125}}$$
$${\tan ^2}x = {1 \over 3}$$
$${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {2 \over {125}}$$
Explanation
It is given that
$${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5}$$
$$3{\sin ^4}x + 2{(1 - {\sin ^2}x)^2} = {6 \over 5}$$
$$ \Rightarrow 25{\sin ^4}x - 20{\sin ^2}x + 4 = 0$$
$$ \Rightarrow {\sin ^2}x = {2 \over 5}$$ and $${\cos ^2}x = {3 \over 5}$$
Hence, $${\tan ^2}x = {2 \over 3}$$
Therefore, $${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {1 \over {125}}$$
Comments (0)
