Question

If 3 cot A=4,check whether 1-tan²A /1+tan²A=cos²A-sin²A or not



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Answer 1

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Answer 2

Answer:

$\huge\rm\red{ANSWER}$

Let us consider a right triangle ABC in which ∠B=90°.

Now,

$\bf{cot \:A = \frac{Base}{Perpendicular } = \frac{AB}{ {BC} } = \frac{4}{3} }$

Let AB=4k and BC=3k

∴ By Pyhthagoras Theorem

AC² = AB² + BC²

AC²=(4k) ² + (3k) ²=16k² + 9k²

AC²=25k²

∴ AC=5k

Therefore,

$\bf{tan A = \frac{Perpendicular}{Base}=\frac{BC}{AB}=\frac{3k}{4k}=\frac{3}{4}}$

and,

$\bf{sin A = \frac{Perpendicular}{Hypotenuse }=\frac{BC}{AC}= \frac{3k}{4k}=\frac{3}{4}}$

$\bf{cos A = \frac{Base}{Hypotenuse }=\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5}}$

Now,

$\bf{LHS =\frac{1–tan² A}{1+tan² A}}$

$\bf{=\frac{1–(\frac{3}{4})²}{1–(\frac{3}{4})²}=\frac{1– \frac{9}{16}}{1+ \frac{9}{16}}= \frac{16–9}{16+9}= \frac{7}{25}}$

$\bf{RHS= cos² A–sin² A=(\frac{4}{5})² –(\frac{3}{5})² = \frac{16}{25} – \frac{9}{25} = \frac{7}{25}}$

Hence,

[tex]\bf{\frac{1–tan²A}{1+tan²A}=cos²A –sin²A}[tex]

[tex]\huge\rm\blue{THANKYOU}[tex]