Question

pls answer correctly ​

Answer 1

Answer:

I have explained in the photo. See this photo.

Answer 2

Answer:

According to your question,

3cotA = 4 and we have to check whether

$\frac{1 - { \tan(a) }^{2} }{1 + { \tan(a) }^{2} } = cos {}^{2} a - {sin}^{2} a$

,

So , let's start from the LHS

3cotA = 4

cotA = 4/3 and we know that

$cot(a) = \frac{1}{ \tan(a) }$

so, tan(A) = 3/4

now putting the value of tan(A) in LHS

$\frac{1 - \tan {}^{2} (a) }{1 + \tan {}^{2} (a) } = \frac{1 - ( \frac{3}{4}) {}^{2} }{1 + ( \frac{3}{4}) {}^{2} } \\ \\ = \geqslant \frac{1 - \frac{9}{16} }{1 + \frac{9}{16} } \\ \\ = \geqslant \frac{ \frac{16 - 9}{16} }{ \frac{16 + 9}{16} } = \frac{ \frac{7}{16} }{ \frac{25}{16} } = \frac{7}{16} \times \frac{16}{25} \\ \\ = \geqslant \frac{7}{25}$

LHS = 7/25

Now , let's find the RHS

$\tan(a) = \frac{ \sin(a) }{ \cos(a) } = \frac{3}{4}$

$\tan(a) = \frac{prependicular}{base} = \frac{3}{4}$

Using Pythagoras theorem

H² = P² + B²

H² = 3² + 4²

H² = 9 + 16

H² = 25

Hypotenuse = 5cm , Prependicular = 3cm and base = 4cm

$\sin(a) = \frac{prependicular}{hypotenuse} = \frac{3}{5} \\ \\ = \geqslant \cos(a) = \frac{base}{hypotenuse} = \frac{4}{5}$

$now \: putting \: the \: value \\ \\ = \geqslant \cos {}^{2} (a) - \sin {}^{2} (a) \\ \\ = \geqslant (\frac{4}{5}) {}^{2} - ( \frac{3}{5} ) {}^{2} \\ \\ = \geqslant \frac{16}{25} - \frac{9}{25} \\ \\ = \geqslant \frac{16 - 9}{25} \\ \\ = \geqslant \frac{7}{25} \\ \\ \\ = \geqslant lhs = rhs(proved)$