If 3 cot θ=4, then frac{1-tan ^{2} θ}{1+tan ^{2} θ}=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

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

$\frac{7}{25}$

$\frac{4}{3}$

$\frac{3}{4}$

$\frac{1}{7}$

Solution

$3 \cot \theta=4\, \therefore \cot \theta=\frac{4}{3}\, \therefore \tan \theta=\frac{3}{4}$

$\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{1-\left(\frac{3}{4}\right)^{2}}{1+\left(\frac{3}{4}\right)^{2}}=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{16-9}{16+9}=\frac{7}{25}$

Standard 10

Mathematics

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$\frac{\cos 50}{\sin 40}+\frac{\sin 15}{\cos 75}=\ldots \ldots \ldots \ldots$

easy

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

Solution

Similar Questions

$\sin ^{2} 1+\sin ^{2} 3+\sin ^{2} 87+\sin ^{2} 89=\ldots \ldots \ldots \ldots$

$\cos \theta=\frac{15}{17},$ then the value of $\operatorname{cosec} \theta+\cot \theta $ is ………

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