Question

If a=sec theta-tan theta and b=sec theta +tan theta,then

Answer 1

If $a=\sec \theta-\tan \theta$ and $b=\sec \theta +\tan \theta$, then $b^2-a^2=4\sec \theta\tan \theta$.

Step-by-step explanation:

We have,

$a=\sec \theta-\tan \theta$                                   ......... (1)

and $b=\sec \theta +\tan \theta$                            ......... (2)  

Squaring (1) and (2), and subtracting them, we get  

$a^2-b^2=(\sec \theta-\tan \theta)^2-(\sec \theta+\tan \theta)^2$  

⇒ $a^2-b^2=(\sec^2 \theta+\tan^2 \theta-2\sec \theta\tan \theta)-(\\sec^\theta+\tan^2 \theta+2\sec \theta\tan \theta)$  

⇒ $a^2-b^2=\sec^2 \theta+\tan^2 \theta-2\sec \theta\tan \theta-\\sec^\theta-\tan^2 \theta-2\sec \theta\tan \theta$  

⇒ $a^2-b^2=-2\sec \theta\tan \theta-2\sec \theta\tan \theta$

⇒ $a^2-b^2=-4\sec \theta\tan \theta$

⇒ $b^2-a^2=4\sec \theta\tan \theta$

If $a=\sec \theta-\tan \theta$ and $b=\sec \theta +\tan \theta$, then $b^2-a^2=4\sec \theta\tan \theta$.

Answer 2

Answer:

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Step-by-step explanation:

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