Prove that 1 + tan 2 A = sec 2 A

Q) Prove that 1 + tan ² A = sec ² A

Ans:

Step 1: Let’s draw a right angled triangle:

Here ABC is a triangle where ∠ B is right angle.

Step 2: By applying Pythagoras theorem, we know:

AB ²+ BC ²= AC ²

Step 3: Let’s divide the above equation by AB on both sides, we get:

$\frac{AB^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$

∴ $\frac{\cancel {AB^2}}{\cancel {AB^2}} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$

∴ $1 + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$

Step 4: Now, by observation, in the above diagram:

$\frac{BC}{AB} = \tan A$

and $\frac{AC}{AB} = \sec A$

Step 5: By substituting 2 values in the above equation, we get:

Existing equation: $1 + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$

∴ 1 + tan ²A = sec ²A

Hence Proved !

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