Question

X=cosec49 then
tan^249+1/sin^249sec^241+sin^241+1/cosec^249​

Answer 1

from the above triangle, if angle C = 49° and angle B = 90°, the we get angle A = 41°

since

cosec49° = x

then, with respect to angle C

perpendicular = 1

hypotenuse = x

base = √(x² - 1) ... (using pythagoras theorem)

$tan {49}^{o} = \frac{1}{ \sqrt{ {x}^{2} - 1} } \\ {tan}^{2} {49}^{o} = \frac{1}{ {x}^{2} - 1}$

similarly,

${sin}^{2} {49}^{o} = \frac{1}{ {x}^{2} } \\ {sec}^{2} {41}^{o} = \frac{ {x}^{2} }{1} = {x}^{2} \\ {sin}^{2} {41}^{o} = \frac{ {x}^{2} - 1 }{ {x}^{2} } = 1 - \frac{1}{ {x}^{2} } \\ {cos}^{2} {41}^{o} = \frac{1}{ {x}^{2} }$

Now,

$\: \: \: \: \: \: \: {tan}^{2} {49}^{o} + \frac{1}{ {sin}^{2} {49}^{o} {sec}^{2} {41}^{o} } \\ \: \: \: \: \: \: + {sin}^{2} {41}^{o} + \frac{1}{ {cosec}^{2} {49}^{o} } \\ = {tan}^{2} {49}^{o} + {cosec}^{2} {49}^{o} {cos}^{2} {41}^{o} \\ \: \: \: \: \: + {sin}^{2} {41}^{o} + {sin}^{2} {49}^{o} \\ = \frac{1}{ {x}^{2} - 1} + ( \frac{1}{ {x}^{2} } )( {x}^{2} ) + 1 - \frac{1}{ {x}^{2} } + \frac{1}{ {x}^{2} } \\ = \frac{1}{ {x}^{2} - 1} + 1 + 1 + 0 \\ = \frac{1}{ {x}^{2} - 1} + 2 \\ = \frac{1 + 2 {x}^{2} - 2 }{ {x}^{2} - 1 } \\ = \frac{2 {x}^{2} - 1}{ {x}^{2} - 1 } \\ = \frac{( {x}^{2} - 1) + {x}^{2} }{ {x}^{2} - 1 } \\ = 1 + \frac{ {x}^{2} }{ {x}^{2} - 1 }$