Question

Solve ${\left\{\frac{3\cos 43}{\sin 47}\right\}}^2-\frac{\cos 37.\csc 53}{\tan 5.\tan 25.\tan 45.\tan 65.\tan 85.}=$

• A. $7$
• B. $9$
• C. $1$
• D. $8$

Answer 1

The correct option is D $8$

${\left\{\frac{3cos43}{sin47}\right\}}^2-\frac{cos37\cdot csc53}{tan5\cdot tan25\cdot tan45\cdot tan65\cdot tan85}=?$

We know $sinA=cos(90-A)$

Consider ${\left\{\frac{3cos43}{sin47}\right\}}^2$

${\left\{\frac{3cos43}{sin47}\right\}}^2={\left\{\frac{3cos43}{cos(90-47)}\right\}}^2$

$={\left\{\frac{3cos43}{cos43}\right\}}^2$

$=3^2$

${\left\{\frac{3cos43}{sin47}\right\}}^2=9$ ....(1)

Consider the numerator $cos37\cdot csc53$

We know $cosecA=sec(90-A)$ and $sec(90-A)=\frac{1}{cosA}$

$cos37\cdot csc53=cos37\cdot sec(90-37)=cos37\cdot \frac{1}{cos37}=1$

Consider the denominator: $tan5\cdot tan25\cdot tan45\cdot tan65\cdot tan85$

we know $tanA=cot(90-A)=\frac{1}{tan(90-A)}$. Apply this for denominator.

$tan5\cdot tan25\cdot tan45\cdot tan65\cdot tan85=tan5\cdot tan85\cdot tan45\cdot tan25\cdot tan65$

$=tan5\cdot \frac{1}{tan5}\cdot tan45\cdot tan25\cdot \frac{1}{25}$

$=tan45$

$tan5\cdot tan25\cdot tan45\cdot tan65\cdot tan85=1$

$\frac{cos37\cdot csc53}{tan5\cdot tan25\cdot tan45\cdot tan65\cdot tan85}=\frac{1}{1}=1$ ....(2)

From $\left(1\right),\left(2\right)$ we get

${\left\{\frac{3cos43}{sin47}\right\}}^2-\frac{cos37\cdot csc53}{tan5\cdot tan25\cdot tan45\cdot tan65\cdot tan85}=9-1=8$