If 3 cot theta = 4,then frac{1 - tan^2 theta} | vedclass

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(A) Given $3 \cot 	heta = 4$,we have $\cot 	heta = \frac{4}{3}$.Since $	an 	heta = \frac{1}{\cot 	heta}$,we get $	an 	heta = \frac{3}{4}$.Now,s

Vedclass · Accepted Answer

$\frac{7}{25}$

Vedclass · Answer

(A) Given $3 \cot 	heta = 4$,we have $\cot 	heta = \frac{4}{3}$.Since $	an 	heta = \frac{1}{\cot 	heta}$,we get $	an 	heta = \frac{3}{4}$.Now,substitute the value of $	an 	heta$ into the expression:$\frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} = \frac{1 - (\frac{3}{4})^2}{1 + (\frac{3}{4})^2}$$= \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}}$$= \frac{\frac{16 - 9}{16}}{\frac{16 + 9}{16}}$$= \frac{7}{25}$.

If $3 \cot \theta = 4$,then $\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \dots$

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