If 3 cot A = 4, check whether \frac{1-tan^2A}{1+tan^2A}=cos^2A-sin^2A or not ?

byAmlesh Mahato •May 30, 2021

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Solution :

⇨ Given

⇨ 3cot A = 4

⇨ cot A = 4/3 = b/p

⇨ Therefore b = 4, p = 3, h = ?

⇨From Pythagoras Theorem

$\large h^2=p^2+b^2$

$\Rightarrow h^2=3^2+4^2$

$\Rightarrow h^2=9+16$

$\Rightarrow h^2=25$

$\Rightarrow h=5$

⇨ L.H.S

$8. If 3 cot A = 4, check whether \frac{1-tan^{2}A}{1+tan^{2}A}=cos^2A-sin^2A or not ?$

$=\frac{1-\left ( \frac{p}{b} \right )^{2}}{1+\left ( \frac{p}{b} \right )^{2}}$

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

$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$

$=\frac{\frac{16-9}{16}}{\frac{16+9}{16}}$

$=\frac{16-9}{16}\times \frac{16}{16+9}$

$=\frac{7}{25}$

⇨ R.H.S

$cos^2A-sin^2A$

$= \left ( \frac{b}{h} \right )^{2}-\left ( \frac{p}{h} \right )^{2}$

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

$=\frac{16}{25}-\frac{9}{25}$

$=\frac{16-9}{25}$

$=\frac{7}{25}$

⇨Therefore L.H.S = R.H.S

Tags: Chapter-8 Exercise-8.1 Math- Class 10

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