Question

Hola Brainlians. Here's a question to check your Trigonometric skill. ICSE Board : Trigonometry, Class 10. Give answer fast and no spam.

Proof the given identity.

sec^2 x = 1 + tan^2 x

Spammers stay away. No spam. Give best answer. Mods and stars answer.​

Answer 1

Answer:

Starting from:

cos2(x)+sin2(x)=1

Divide both sides by cos2(x) to get:

cos2(x)cos2(x)+sin2(x)cos2(x)=1cos2(x)

which simplifies to:

1+tan2(x)=sec2(x)

Answer 2

Answer:

Step-by-step explanation:

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

For this, we will first simplify RHS to get back LHS.

Here, we go towards right, the right path:-

We know that:-

$\sf{tanx = \dfrac{sinx}{cosx}}$----(1)

And, we also know this one :-

$\sf{secx = \dfrac{1}{cosx}}$ ------(2)

Let's start the mechanism to get the resultant.  

$\sf{1 + \dfrac{sin^2 x}{ cos^2 x }}$

$\sf{\dfrac{cos^2 x + sin^2 x}{ cos^2 x }}$

 

We know that $\sf{sin^2 x + cos^2 x = 1}$

Applying this identity, we get

$\sf{\dfrac{1}{ cos^2 x }}$

Let's move towards left side, LHS.

As said above in (2), so we can now substitute the value of secx to cosx.

$\implies \sf{\dfrac{1}{ cos^2 x }}$

Now, compare LHS and RHS, it is same. So, we can say that $\sf{sec^2 x = 1+ tan^2 x}$

Hence Proved!