Question

Prove that (1+tanx)2 +(1+cotx)2=(secx+cosecx)2

Answer 1

this is proof of this questions.....

I hope you understand thiis question..

Answer 2

$\huge\mathfrak{Answer:}$

$\large\underline{\sf{\blue{Given:}}}$

• We have been given a trigonometric expression
• Containing some trigonometric functions

$\large\underline{\sf{\blue{To \: Prove:}}}$

$\boxed{\sf{(1 + tan \: x)^2 + (1+cot \: x)^2 = (sec \: x + cosec \: x)^2}}$

$\large\underline{\sf{\blue{Proof:}}}$

We have been given a trigonometric equation

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$\large\underline{\mathfrak{\red{According \: to \: the \: Question:}}}$

Taking Left Hand Side of the Equation and proving it equal to Right Side

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$\sf{(1 + tan \: x)^2 + (1+cot \: x)^2}$

$\sf{1 + tan^2 \: x + 2 tan \: x + 1+cot^2 \: x + 2 cot \: x}$

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Using algebric identity in above equation

$\boxed{\sf{\green{(a+b)^2 = a^2 + b^2 + 2ab}}}$

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$\sf{1 + tan^2 \: x + 1 + cot^2 \: x + 2 tan \: x + 2 cot \: x}$

$\sf{sec^2 \: x + cosec^2 \: x + 2 ( tan \: x + cot \: x)}$

$\sf{sec^2 \: x + cosec^2 \: x + 2 ( tan \: x + cot \: x)}$

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Converting tangent and cotangent in terms of sine and cosine component

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$\sf{sec^2 \: x + cosec^2 \: x + 2 \left ( \dfrac{sin \: x}{cos \: x}+ \dfrac{cos \: x}{sin \: x } \right )}$

$\sf{sec^2 \: x + cosec^2 \: x + 2 \left ( \dfrac{sin^2 \: x + cos^2 \: x}{sin \: x \: cos \: x} \right )}$

$\sf{}$

Using Trigonometric Equation below

$\boxed{\sf{\green{sin^2x + cos^2x = 1}}}$

$\sf{}$

$\sf{sec^2 \: x + cosec^2 \: x + 2 \left ( \dfrac{1}{sin \: x \: cos \: x} \right )}$

$\sf{sec^2 \: x + cosec^2 \: x + 2 sec \: x \: cosec \: x }$

$\sf{sec^2 \: x + cosec^2 \: x + 2 sec \: x \: cosec \: x }$

$\sf{(sec \: x + cosec \: x)^2}$

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$\large{\purple{\underline{\underline{\mathtt{Trigonometric \: Identities :}}}}}$

• sin²x + cos²x = 1
• sec²x - tan²x = 1
• cosec²x - cot²x = 1
• secx = 1 / cosx
• cosecx = 1 / sinx
• tanx = 1 / cot x

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