Question

if sec alfa is 5÷4 verify that tan alfa÷1+tan^2 alfa=sin alfa ÷sec alfa​

Answer 1

Solution:

Given that ;

$\sec \alpha = \dfrac{5}{4}$

We know that ;

$\mathrm{ \sec \alpha = \dfrac{hypotense}{base}} = \dfrac{5}{4}$

So,

• Hypotenuse = 5
• Base = 4

We can calculate perpendicular, using the Pythagoras theorem -

$\mathrm{perpendicular = \sqrt{h {}^{2} - b {}^{2} } } \\ \\ \mathrm{perpendicular = \sqrt{(5) {}^{2} - (4) {}^{2}}} \\ \\ \mathrm{perpendicular = \sqrt{25 - 16 } } \\ \\ \mathrm{perpendicular = \sqrt{9 } } \\ \\ \mathrm{perpendicular = 3}$

Now,

$\rm{ \sin \alpha = \frac{p}{h}} = \frac{3}{5} \\ \\ \rm{ \tan \alpha = \frac{p}{b} = \frac{3}{4} }$

We have to verify that,

$\rm{ \frac{ \tan \alpha}{1 + \tan ^{2} \alpha } } = \frac{ \sin \alpha }{ \sec \alpha } \\ \\ \frac{ \frac{3}{4} }{1 + \left( \frac{3}{4} \right)^{2}} = \frac{ \frac{3}{5} }{ \frac{5}{4} } \\ \\ \frac{3}{4} \div \left(1 + \frac{9}{16} \right) = \frac{3}{5} \div \frac{5}{4} \\ \\ \frac{3}{4} \div \left( \frac{16 + 9}{16} \right) = \frac{3}{5} \times \frac{4}{5} \\ \\ \frac{3}{4} \times \frac{16}{25} = \frac{12}{25} \\ \\ \frac{12}{25} = \frac{12}{25}$

Hence, it is verified.