Question

Tan a + sin a / tan a - sin a = sec a +1/ sec a -1

Answer 1

Answer:

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Answer 2

Given: Here we have given $\dfrac{tan a +sin a}{tan a- sin a } =\dfrac{(sec a +1)}{(sec a -1)}$

To find: we have to prove LHS=RHS

Solution:

• we will take the LHS first

$\dfrac{tan a +sin a}{tan a- sin a } =\dfrac{\dfrac{sin a}{cos a } +sin a}{\dfrac{sin a}{cos a } - sin a }$

• by taking LCM we get,

$=\dfrac{\dfrac{sin a+sin acos a}{cos a } }{\dfrac{sin a-sin acos a}{cos a } }\\\\=\dfrac{sin a+sin acos a}{sin a-sin acos a }$

• Now we will take Sina in common

$=\dfrac{sin a(1+cos a)}{sin a(1-cos a)}\\\\=\dfrac{(1+cos a)}{(1-cos a)}$

• we know that $cosa=\dfrac{1}{seca}$

$=\dfrac{(1+\dfrac{1}{seca} )}{(1-\dfrac{1}{seca})}\\\\=\dfrac{(\dfrac{seca+1}{seca} )}{(\dfrac{seca-1}{seca})}$

$=\dfrac{(sec a +1)}{(sec a -1)} =RHS$

Final answer:

Hence RHS=LHS