Question

If 3 cot A = 4, check whether (1 – tan²A)/(1 + tan²A) = (cos²A — sin²A) or not.

Answer 1

Step-by-step explanation:

From question the condition given is:

$\bold{3 \cot A = 4}$

$\bold{\cot A=\frac{4}{3}}$

$\bold{\Rightarrow \tan A=\frac{3}{4}}$

LHS:

$\bold{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}}$

$\bold{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}}$

$\bold{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{\frac{16-9}{16}}{\frac{16-9}{16}}}$

$\bold{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{7 / 16}{25 / 16}}$

$\bold{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{7}{16} \times \frac{16}{25}}$

$\bold{\therefore \frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\frac{7}{25}}$

Now, RHS:

$\bold{\sin A=\frac{\text { Opposite side }}{\text { Hypotenuse }}}$

$\bold{\cos A= \frac{\text { Adjacent side }}{\text { Hypotenuse }}}$

$\bold{\tan A=\frac{\text { Opposite side }}{\text { Adjacent side }}}$

$\bold{Hypotenus = \sqrt{(Opposite \ side)^2+(Adjacent \ side)^2}}$

$\bold{Hypotenus = \sqrt{(3)^2+(4)^2}}$

$\bold{Hypotenus = \sqrt{9+16} = \sqrt{25}=5}$

$\bold{\therefore \sin A=\frac{3}{5}}$

$\bold{\therefore \cos A=\frac{4}{5}}$

$\bold{\cos ^{2} A-\sin ^{2} A=\left(\frac{4}{5}\right)^{2}-\left(\frac{3}{5}\right)^{2}}$

$\bold{\cos ^{2} A-\sin ^{2} A=\frac{16}{25}-\frac{9}{25}}$

$\bold{\therefore \cos ^{2} A-\sin ^{2} A=\frac{7}{25}}$

Thus,

$\bold{\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=\cos ^{2} A-\sin ^{2} A}$

Hence proved.

Answer 2

Answer

hey mate your answer is in attachment...

Step by step explanation.

Hope it helps!☺️