Math, asked by FlashNish, 1 year ago

TRIGONOMETRY. only step by step answer no fake answer plz ​

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Answered by shadowsabers03
4

     

Hey mate!!!

Here's the proof!!!

 

\Rightarrow\ \boxed{LHS} \\ \\ \Rightarrow\ \boxed{\frac{\cot^2 A(\sec A-1)}{1+\sin A}+\frac{\sec^2A(\sin A-1)}{1+\sec A}} \\ \\ \\ \Rightarrow\ \boxed{\frac{\cot^2A(\sec A-1)(\sec A+1)+\sec^2A(\sin A-1)(\sin A+1)}{(1+\sec A)(1+\sin A)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{\cot^2A(\sec^2 A-1)+\sec^2A(\sin^2 A-1)}{(1+\sec A)(1+\sin A)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{\cot^2 A \cdot \tan^2A+\sec^2A \cdot \sin^2A-\sec^2A}{(1+\sec A)(1+\sin A)}}

\Rightarrow\ \boxed{\frac{1+\tan^2A-\sec^2A}{(1+\sin A)(1+\sec A)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{1+(-1)}{(1+\sin A)(1+\sec A)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{0}{(1+\sin A)(1+\sec A)}} \\ \\ \\ \Rightarrow\ \boxed{0} \\ \\ \Rightarrow\ \boxed{RHS}

Hence proved!!!

Plz ask me if you've any doubt on my answer.

Thank you......

\mathfrak{\#adithyasajeevan}

     


FlashNish: hi
FlashNish: can you please explain me 5th step
shadowsabers03: Okay.

We know that sec^2 - tan^2 = 1, so that, tan^2 = sec^2 - 1. So in the numerator, sec^2 - 1 which is included in the bracket becomes tan^2 and will be multiplied with cot^2 at left to become 1, because both are reciprocals to each other.

Then, in sec^2(sin^2-1), sec^2 is multiplied inside sin^2-1 (distributive law) and thus we get tan^2-sec^2. So the numerator becomes 1 + tan^2 - sec^2.
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