Question

1-tan square theta/1+tan square theta =(cos square theta - sin square theta)​

Answer 1

$\boxed{\dfrac{1-tan^{2}\theta}{1+tan^{2}\theta}=cos^{2}\theta-sin^{2}\theta}$

Concept to be used:

• $tan\theta=\dfrac{sin\theta}{cos\theta}$

• $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b}\times\dfrac{d}{c}$

• $sin^{2}\theta+cos^{2}\theta=1$

• $1+tan^{2}\theta=sec^{2}\theta$

• $\dfrac{a\pm b}{c}=\dfrac{a}{c}\pm\dfrac{b}{c}$

• $cos^{2}\theta\:sec^{2}\theta=1$

Step-by-step explanation:

Method 1.

Here, L.H.S. is

$\dfrac{1-tan^{2}\theta}{1+tan^{2}\theta}$

$=\dfrac{1-\dfrac{sin^{2}\theta}{cos^{2}\theta}}{1+\dfrac{sin^{2}\theta}{cos^{2}\theta}}$

$=\dfrac{\dfrac{cos^{2}\theta-sin^{2}\theta}{cos^{2}\theta}}{\dfrac{cos^{2}\theta+sin^{2}\theta}{cos^{2}\theta}}$

$=\dfrac{cos^{2}\theta-sin^{2}\theta}{cos^{2}\theta}\times\dfrac{cos^{2}\theta}{cos^{2}\theta+sin^{2}\theta}$

$=\dfrac{cos^{2}\theta-sin^{2}\theta}{1}\times\dfrac{1}{1}$

$=cos^{2}\theta-sin^{2}\theta$

that is, R.H.S.

Thus proved.

Method 2.

Here, L.H.S. is

$\dfrac{1-tan^{2}\theta}{1+tan^{2}\theta}$

$=\dfrac{1-tan^{2}\theta}{sec^{2}\theta}$

$=\dfrac{1}{sec^{2}\theta}-\dfrac{tan^{2}\theta}{sec^{2}\theta}$

$=cos^{2}\theta-\dfrac{sin^{2}\theta}{cos^{2}\theta}\dfrac{1}{sec^{2}\theta}$

$=cos^{2}\theta-sin^{2}\theta$

that is, R.H.S.

Thus proved.

NOTE:

However,

$\dfrac{1-tan^{2}\theta}{1+tan^{2}\theta}=cos^{2}\theta-sin^{2}\theta$

$\Rightarrow \dfrac{1-tan^{2}\theta}{1+tan^{2}\theta}=cos2\theta$

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