Math, asked by KarthikVatsavayi, 1 year ago

tan A - 1 + SEC a/ tan a + 1 - sec a

Answers

Answered by Swarnimkumar22

$\bold{Hay Dear Friend }$

$\frac{tan \: a - 1 + sec \: a}{tan \: a + 1 - sec \: a}$

$\frac{(tan \: a + sec \: a) - ( - tan {}^{2}a + {sec}^{2}a) }{tan \: a + 1 - sec \: a}$

$\italic{we know that formula a^2-b^2=(a-b) (a+b) so, put it value in given equation }$

$\frac{tan \: a + sec \: a - (sec \: a + tan \: a)(sec \: a - tan \: a)}{tan \: a + 1 - sec \: a}$

$\frac{(tan \: a + sec \: a)(1 - sec \: a + tan \: a)}{tan \: a + 1 - sec \: a}$

$\frac{tan \: a - tan \: a \: sec \: a + {tan}^{2}a + sec \: a - sec {}^{2} a + sec \: a \: tan \: a }{tan \: a + 1 - sec \: a}$

$\Italic{Now, we get that }$

secA+tanA

$\italic{HENCE SOLVED }$

$\color {Blue}{Swarnimkumar21}$

Swarnimkumar22: :-)

KarthikVatsavayi: Sir, how did you get SEC A+ TAN A from the previous step.

Swarnimkumar22: Just solve that we found many equations are cut

Swarnimkumar22: see carefully I hope you understand

Swarnimkumar22: :-)

Answered by shahjignesh158

Answer:

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