Question

please answer my question
$tan ^{2} 60 + 4 \cos^{2} 45 + 3sec ^{2} 30 + 5cos^{2} 90 \div \csc(30) + \sec(60) - \cot ^{2}30$
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Answer 1

Solution:-

$\frac{{tan}^{2} 60\degree + 4 \: {cos}^{2} 45\degree + 3 \: {sec}^{2} 30\degree + \: {cos }^{2} 90\degree}{cosec \: 30\degree+ sec \: 60\degree - {cot}^{2}30\degree}$

Tan 60° = √3

Cos 45° = 1/√2

Sec 30° = 2/√3

Cos 90° = 0

Cosec 30° = 2

Sec 60° = 2

Cot 30° = √3

So,

$\implies \: \frac{ ({ \sqrt{3}) }^{2} + 4 \times {( \frac{1}{ \sqrt{2} } ) }^{2} + 3 \times {( \frac{2}{ \sqrt{3} } )}^{2} + {(0)}^{2}}{2 + 2 - {(3)}^{2} } \\\\ \implies \: \frac{3 + 4 \times \frac{1}{2} + 3 \times \frac{4}{3} }{4 - 3}\\ \\ \implies \frac{3 + 2 + 4}{1} \\ \\ \implies9$

Therefore,

$\frac{{tan}^{2} 60\degree + 4 \: {cos}^{2} 45\degree + 3 \: {sec}^{2} 30\degree + \: {cos }^{2} 90\degree}{cosec \: 30\degree + sec \: 60\degree - {cot}^{2}30\degree }$ = 9.