Question

prove that 1 + cot²θ = cosec²θ​

Answer 1

sin²a+cos²a=1

1+cot²a=1/sin²a [division with sin²a]

1+cos²=cosec²a

proved

hope this is helpful

Answer 2

$\large\bold{\underline{\underline{\blue{\sf solution:-}}}}$

imagine a right angled triangle ABC.

since angle B is 90°, we can apply Pythagoras theorem.

$\tt\implies AC^2=AB^2+BC^2$

$\red{\tt divide \: by \: AC^2}$

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

$\boxed{\tt 1=sin^2A + cos^2A}$

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$\boxed{\green{\sf{divide\:by\:sin^2A}}}$

$\tt \dfrac{1}{sin^2A}=\dfrac{sin^2A}{sin^2A}+\dfrac{cos^2A}{sin^2A}$

$\tt cosec^2A = 1 + cot ^2A$

$\boxed{\red{\tt hence\:proved}}$

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$\huge\bigstar\bold{\pink{\sf concepts \: used}}\bigstar$

$\tt \implies \dfrac{1}{sin^2A}=cosec^2A$

$\tt \implies \dfrac{cos^2A}{sin^2A}=cot^2A$

$\tt \implies \dfrac{BC^2}{AC^2}=sin^2A$

$\tt \implies \dfrac{AB^2}{AC^2}=cos^2A$

basically we proved that 1=sin²A+cos²A and then divided the equation with sin²A to derive an equation with cot and cosec.