Question

prove that 1 +sec a-tan a/ 1+sec a + tana= 1+ sin a / cos a

Answer 1

Here is ur solution

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Answer 2

The given trigonometric equation is proved using the trigonometric and algebraic identities.

( The RHS must have '-' instead of '+' )

We need to prove the trigonometric equation i.e. show LHS = RHS :

        $\frac{1+secA- tanA}{1+secA+ tanA} = \frac{1-sinA}{cosA}$

We begin solving the LHS and by simplifying bring it equal to LHS to prove the given equation.

Step 1 : Using the Square Trigonometric Identity

We use the trigonometric identity between the squares of the sec and tan ratios. The identity is :

                     $sec^{2}x = 1 + tan^{2}x$

So, we replace the 1 in the numerator with : $sec^{2}x - tan^{2}x$

             $= \frac{1+secA- tanA}{1+secA+ tanA}$

             $= \frac{sec^{2}A -tan^{2}A +secA- tanA}{1+secA+ tanA}$

Step 2 : Splitting the algebraic square identity

We use the algebraic identity : $a^{2} -b^{2} =(a+b)(a-b)$

             $= \frac{(secA -tanA)(secA+tanA) +secA- tanA}{1+secA+ tanA}$

             $= \frac{(secA -tanA)(secA+tanA) +(secA- tanA)}{1+secA+ tanA}$

             $= \frac{(secA -tanA)(secA+tanA+1)}{1+secA+ tanA}$

             $=(secA -tanA)$

Step 3 : Converting the trigonometric rations sec and tan to base ratios i.e. cos and sin :

                           $secA = 1/cosA\\tanA = sinA/ cosA$

Thus, replacing the ratios, we get :

           $= (1/cosA) - (sinA/cosA)$

           $=(1-sinA)/cosA$

           $= RHS$

Hence, the given trigonometric equation is proved.

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