Question

Solve it pleaseeeeeee............fast.......

Answer 1

$\blue{\bold{answer = 8}}$

$= = = = = = = = = = = = = = = = = = =$

_______________________________________________________

Explanation:-

• Given cosec A= √2

$\red{cosec \: theta = \frac{hypotenus(h)}{perpendicular(p)} }$

N

$= > \frac{h}{p} = \sqrt{2} \\ \\ = > \red{ \boxed{h = \sqrt{2} \: and \: p = 1 }}$

Now ,by applying Pythagoras theorem we can find the base of the triangle ✓ which is as follows✓

$= > \green{b = \sqrt{h^{2} - p ^{2} } } \\ \\ = > \green{b = \sqrt{ (\sqrt{2})^{2} - 1^{2} } } \\ \\ = > \green{b = \sqrt{2 - 1} } \\ \\ = > \green{b = \sqrt{1} = 1 }$

So by the above calculation we find the base of the triangle is

$\red{1}$

Now ,we have to find the other trigonometric ratios ✓

$sin \: a = \frac{p}{h} = \frac{1}{ \sqrt{2} } \\ \\ = > cot \: a = \frac{b}{p} = \frac{1}{1} = 1 \\ \\ = > tan \: a = \frac{p}{b} = \frac{1}{1} = 1 \\ \\ = > cos \: a = \frac{b}{h} = \frac{1}{ \sqrt{2} }$

Now let us solve this problem by using the ratios ✓

$= > \frac{2sin^{2}a + 3cot^{2} a}{tan ^{2}a - cos ^{2}a} \\ \\ = > \frac{2 \times ( \frac{1}{ \sqrt{2} } ) ^{2} + 3 \times 1 ^{2} }{1^{2} - (\frac{1}{ \sqrt{2} }) ^{2} } \\ \\ = > \frac{ \frac{2}{2} + 3}{1 - \frac{1}{2} } \\ \\ = > \frac{1 + 3}{ \frac{2 - 1}{2} } \\ \\ = > \frac{4}{ \frac{1}{2} } \\ \\ = > \red{8}$