Question

Evaluate
cosec^2(90-theta)-tan^2theta/5(cos^2 48+cos^42)+2/5sin48sec42-1/5tan^60​

Answer 1

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★ CORRECT QUESTION ★

$\frac{ {cosec }^{2}(90 - x) - {tan}^{2} x}{5( {cos}^{2} 48 + {cos}^{2}42) } + \frac{2}{ 5} \times sin \: 48 \: sec \: 42 - \frac{1}{5} {tan}^{2} 60$

★ SOLUTION ★

ACCORDING TO THE QUESTION

( LET ∅ = X )

$\frac{ {cosec }^{2}(90 - x) - {tan}^{2} x}{5( {cos}^{2} 48 + {cos}^{2}42) } + \frac{2}{ 5} \times sin \: 48 \: sec \: 42 - \frac{1}{5} {tan}^{2} 60$

NOW COMPLEMENTARY ANGLES OF TRIGNOMATREY:-

$cosec \: (90 - x) = sec \: x \\ \\ sin \: (90 - x) = cos \: x$

$\frac{ {sec }^{2}x - {tan}^{2} x}{5( {cos}^{2} 48 + {sin}^{2}48) } + \frac{2}{ 5} \times sin \: 48 \: cosec \: 48 - \frac{1}{5} {tan}^{2} 60$

NOW BY TRIGNOMATRIC IDENTITY

${sec}^{2} x - {tan}^{2} x = 1 \\ \\ {sin}^{2} x \: + {cos}^{2} = 1$

$\frac{1}{5( 1) } + \frac{2}{ 5} \times sin \: 48 \: cosec \: 48 - \frac{1}{5} {tan}^{2} 60$

NOW BY TRIGNOMATRIC RATIOS

$sin \: x \times cosec \: x = 1$

$\frac{1}{5 } + \frac{2}{ 5} \times1- \frac{1}{5} {tan}^{2} 60$

NOW BY TRIGNOMATRIC VALUES

$tan 60 = \sqrt{3}$

$\frac{1}{5 } + \frac{2}{ 5} - \frac{1}{5} { (\sqrt{3} )}^{2}$

NOW BY FURTHER CALCULATION

$= \frac{1}{5 } + \frac{2}{ 5} - \frac{3}{5} \\ \\ = \frac{3}{5} - \frac{3}{5} \\ \\ = 0$

THANKS FOR UR QUESTION HOPE IT HELPS

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