Question

what is the value of
$6tan ^{2} - \frac{6}{cos {}^{2} }$
​

Answer 1

Step-by-step explanation:

$\huge \bull \underline \boldsymbol{ \green{Answer:-}}$

$\implies \bf \red{6 \: \: {tan}^{2} - \dfrac{6}{ {cos}^{2} } }$

we know that,

tan A = sin A / cos A

now,

$\implies \bf \red{6 \: \: \dfrac{ {sin}^{2} }{ {cos}^{2} } - \dfrac{6}{ {cos}^{2} } }$

$\implies \bf \red{ \dfrac{6 \: {sin}^{2} - 6 }{ {cos}^{2} } }$

$\implies \bf \red{ \dfrac{6( {sin}^{2} - 1) }{ {cos}^{2} } }$

we know that,

• sin ²-1 = cos ²

$\implies \bf \red{ \dfrac{6 \: {cos}^{2} }{ {cos}^{2} } }$

$\implies \bf \red{6}$

$\therefore\boldsymbol\green{6 \: \: {tan}^{2} - \dfrac{6}{ {cos}^{2} } = 6}$