Question

ans plz fast explain​

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

To Prove : $\implies \dfrac{cos^2 \alpha - cos^2 \beta}{cos^2 \alpha.cos^2 \beta}=tan^2 \beta - tan^2 \alpha$

Solving left hand side of the given equation :

$\implies \dfrac{cos^2 \alpha - cos^2 \beta}{cos^2 \alpha.cos^2 \beta}$

Adding and subtracting $cos^2 \alpha . cos^2 \beta$ on the numerator.

$\implies \dfrac{cos^2 \alpha - cos^2 \beta+cos^2 \alpha .cos^2 \beta-cos^2 \alpha.cos^2 \beta}{cos^2 \alpha.cos^2 \beta}\\\\\\\implies \dfrac{cos^2 \alpha-cos^2 \alpha.cos^2 \beta - cos^2 \beta+cos^2 \alpha .cos^2 \beta}{cos^2 \alpha.cos^2 \beta}\\\\\\\implies \dfrac{cos^2 \alpha (1 -cos^2 \beta) - cos^2 \beta ( 1 - cos^2\alpha)}{cos^2 \alpha .cos^2\beta}$

From the properties of trigonometry,

1 - cos^2 A = sin^2 A

$\implies \dfrac{cos^2 \alpha sin^2 \beta - cos^2 \beta sin^2\alpha}{cos^2 \alpha .cos^2\beta}\\\\\\\implies \dfrac{cos^2 \alpha sin^2\beta}{cos^2 \alpha . cos^2 \beta}-\dfrac{cos^2\beta sin^2 \alpha}{cos^2\alpha .cos^2\beta}\\\\\\\implies \dfrac{sin^2 \beta}{cos^2 \beta}-\dfrac{sin^2 \alpha}{cos^2 \alpha }\\\\\\\implies tan^2 \beta - tan^2 \alpha$

Hence, proved that  $\implies \dfrac{cos^2 \alpha - cos^2 \beta}{cos^2 \alpha.cos^2 \beta}$=tan^2 \beta - tan^2 \alpha[/tex]