Question

Prove that 1 + tan2A.tanA = sec2A

Answer 1

Given:  1 + tan2A.tanA = sec2A

To find: Prove the above trigonometry.

Solution:

• As we have given the trigonometric terms, so lets consider LHS:

• LHS = 1 + tan2A.tanA
• Solve this further we get:

               1 + (2tanA/1- tan²A) x tanA............     ( tan2A = (2tanA/1- tan²A))

               1 + (2tan²A/1 - tan²A)

               ( 1 - tan²A + 2tan²A/ 1 - tan²A )

               (tan²A + 1/ 1 - tan²A)

               1+ ( tan²A+1 / 1−tan²A) −1

               (1−tan²A+tan²A+1/1−tan²A) −1                  

               (2/1−tan²A) −1                    

               (2tanA×cotA/1−tan²A) −1   .................. (   tan2A = 2tanA×cotA)

               (tan2A×cotA) −1  

• Now, solving further we get:

               ( 2tanA / 1−tan²A )  x ( 1 / tanA ) −1                    

               ( 2 / 1−tan²A ) −1                    

               (2−1+tan²A / 1−tan²A )                

               ( tan²A+1 / 1−tan²A )                

               ( sec²A / 1−tan²A )      

• Convert sec and tan in terms of sin and cos, we get:          

               ( 1 / cos²A / 1−(sin²A/cos²A )   )                

               1 / (cos²A−sin²A)                    

               1 / cos2A  = sec2A.................................RHS

• 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.

Answer:

                    Hence, by above properties, 1 + tan2A.tanA = sec2A       ........hence proved.

Answer 2

Answer:

Step-by-step explanation: