Question

1 + tan 2A.tan A = sec 2A​

Answer 1

Answer:

So by using the formula of tan2A and converting tan and sec in terms of sin and cos, we can prove that 1 + tan2A. tanA = sec2A.

Step-by-step explanation:

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Answer 2

Given:

1 + tan 2A.tan A

To Prove:  

1 + tan 2A.tan A = sec 2A​

We need to Prove

Step-by-step explanation:

The proof is shown as follows:

• Taking Left hand side equation

            $1+tan2A.tanA$

• As we know from trigonometry that

            $tan2A= 2\times tanA /(1-tan^2A)$

• Putting value of $tan2A$ from step 2 into step 1 we get

           $1+tan2A.tanA=1+(2\times tan^2A/(1-tan^2A) ) \\\\ =(1-tan^2A + 2\times tan^2A)/(1-tan^2A)\\\\ =(1+tan^2A)/(1-tan^2A)$

• Now converting tan into cos and sin

        As we can write $tanA=sinA/cosA$

         So now putting value of $tanA$ in the last equation of Step 3

            $1+tan2A.tanA=(1+(sin^2A/cos^2A))/(1-(sin^2A/cos^2A))\\ =(cos^2A + sin^2A)/(cos^2A - sin^2A) \times cosA/cosA$

             As $cos^2A+sin^A=1$, So now our Left hand side equation is

             $1/(cos^2A-Sin^2A)$

• As from trigonometry we can write

               $cos2A=cos^2A-sin^2A$  

• Now putting the value of step 5 equation in the final equation of      left hand side we get

             $1+tan2A.tanA=1/cos2A$

             as $sec2A=1/cos2A$

            so,

            $1+tan2A.tanA=sec2A$

Proved: $1+tan2A.tanA=secA$