Question

Prove 1+tan² A= Sec² A​

Answer 1

Answer:

Step 1: Break 1−sin4Aas a2−b2form.

1−

Answered 5 years ago

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Answer 2

Given -

• Prove 1+tan² A= Sec² A

Solution :

Let us draw a right angle Triangle.

By using Pythagoras Theorem

$\tt\longrightarrow{AB^{2} + AC^{2} = BC^{2}}$

Now, by dividing the each term of both left and side right by (AC)² , we get :

$:\implies\sf{\dfrac{AB^{2}}{AC^{2}} + \dfrac{AC^{2}}{AC^{2}} = \dfrac{BC^{2}}{AC^{2}}}$

$:\implies\sf{\bigg(\dfrac{AB}{AC}\bigg)^{2} + \bigg( \dfrac{AC}{AC}\bigg)^{2} = \bigg( \dfrac{BC}{AC}\bigg)^{2}}$

So, Basically now we know some basic ratios

$\implies\sf{Tan A = \dfrac{Perpendicular}{Base}}$

$:\implies\bf{Tan A = \dfrac{AB}{AC}^{2}}$

Here,

" BC " = Hypotenuse

" AB " = Opposite side ( Perpendicular )

" AC " = Adjacent side ( Base )

$\implies\sf{Sec A = \dfrac{Hypotenuse}{Base}}$

$:\implies\bf{Sec A = \dfrac{BC}{AC}^{2}}$

Thus, after comparing it with the above equation , we get :

$\implies\sf{tan^{2}A + 1 = Sec^{2}A}$

$\implies\sf{ 1 + tan^{2}A = Sec^{2}A}$

__________________________

Method 2 :

As we know that,

$\implies\sf{Sin^{2}A + Cos^{2}A = 1}$

Dividing all the terms by Cos²A , we get :

$:\implies\sf{\dfrac{Sin^{2}}{Cos^{2}} + \dfrac{Cos^{2}}{Cos^{2}} = \dfrac{1^{2}}{Cos^{2}}}$

$:\implies\sf{\bigg(\dfrac{Sin}{Cos}\bigg)^{2} + \bigg( \dfrac{Cos}{Cos}\bigg)^{2} = \bigg( \dfrac{1}{Cos}\bigg)^{2}}$

Some Basic Ratios :

$\implies\sf{Tan A = \dfrac{Sin}{Cos}}$

$\implies\sf{Sec A = \dfrac{1}{CosA}}$

Thus, comparing

Putting values in it, we get :-

$\boxed{\therefore{\sf{tan^{2}A + 1 = Sec^{2}A}}}$

$\boxed{\therefore{\sf{ 1 + tan^{2}A = Sec^{2}A}}}$

Hence , Proved !!