Question

plz answer the question ​

Answer 1

$\huge \green{\mathfrak{solution}}$

$\frac{ {sec}^{2}(90 - \theta) - {cot}^{2} \theta }{2( {sin}^{2} 45 + {sin}^{2} 45) } + \frac{2 {sin}^{2}30 \: {tan}^{2} 38 \: {tan}^{2} 52 }{3( {sec}^{2} 43 - {cot}^{2} \: 47) } \\ \implies \frac{ {cosec}^{2} \theta - {cot}^{2} \theta }{2( {sin}^{2} 45 + {sin}^{2}(90 - 45) } + \frac{2 {sin}^{2} 30 {tan}^{2}(90 - 52) {tan}^{2}52 }{3( {sec}^{2}(90 - 47) - {cot}^{2} 47 } \\ \implies \: \frac{1}{2( {sin}^{2}45 + {cos}^{2}45) } + \frac{2 {sin}^{2} 30 \: {cot}^{2} 52 \: {tan}^{2}52 }{3( {cosec}^{2} 47 - {cot}^{2} 47)} \\ \implies \frac{1}{2(1)} + \frac{2 \times ( \frac{ {1}^{2} }{ {2}^{2} } )}{3(1)} \\ \\ \implies \frac{1}{2} + \frac{1}{6} \\ \\ \implies \frac{3 + 1}{6} \\ \\ \implies \frac{4}{6} = \frac{2}{3}$

Formula

• sin (90-@)=cos @

• sin^2@+cos^@=1

•csc^2@-cot^2@=1

•sec(90-@)=csc@