Hola Brainlians. Here’s a question to check your Trigonometric skill. ICSE Board : Trigonometry, Class 10. Give answer fast and n

2 thoughts on “Hola Brainlians. Here’s a question to check your Trigonometric skill. ICSE Board : Trigonometry, Class 10. Give answer fast and n”

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)

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2. Answer:

   Step-by-step explanation:

   [tex]\sf{sec^2 x = 1+ tan^2 x}[/tex]

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

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

   We know that:-

   [tex]\sf{tanx = \dfrac{sinx}{cosx}}[/tex]—-(1)

   And, we also know this one :-

   [tex]\sf{secx = \dfrac{1}{cosx}}[/tex] ——(2)

   Let’s start the mechanism to get the resultant.

   [tex]\sf{1 + \dfrac{sin^2 x}{ cos^2 x }}[/tex]

   [tex]\sf{\dfrac{cos^2 x + sin^2 x}{ cos^2 x }}[/tex]

   We know that [tex]\sf{sin^2 x + cos^2 x = 1}[/tex]

   Applying this identity, we get

   [tex]\sf{\dfrac{1}{ cos^2 x }}[/tex]

   Let’s move towards left side, LHS.

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

   [tex] \implies \sf{\dfrac{1}{ cos^2 x }}[/tex]

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

   Hence Proved!

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