Question

prove the following identity:​

Answer 1

Question:

Prove-:

$\frac{sec²A}{tan²A}$+$\frac{cosec²A}{cot²A}$=$sec²A×cosec²A$

Answer:

$LHS$=$\large\frac{sec²A}{tan²A}$+$\large\frac{cosec²A}{cot²A}$

$\implies$ $\large\frac{1/cos²A}{sin²A/cos²A}$ + $\large\frac{1/sin²A}{cos²A/sin²A}$

$\implies$ $\large\frac{cos²A}{cos²A×Sin²A}$ + $\large\frac{sin²A}{sin²A×cos²A}$

$\implies$ $\large\frac{1}{sin²A}$×$\large\frac{1}{cos²A}$

$\implies$ $\large\frac{1}{sin²A×cos²A}$ $\: \: \:$ $(1)$

$RHS$=$sec²A×cosec²A$

$\implies$ $\large\frac{1}{cos²A}$×$\large\frac{1}{sin²A}$

$\implies$ $\large\frac{1}{cos²A×sin²A}$ $\: \: \:$ $(2)$

From $(1)$&$(2)$ we get,

$\boxed{LHS=RHS}$

i.e,

$\large\frac{sec²A}{tan²A}$+$\large\frac{cosec²A}{cot²A}$=${sec²A×cosec²A}$

$\underline\color{purple}\boxed{Hence \: Proved!!}$

Answer 2

Answer:

Hope this helps you :)

Step-by-step explanation:

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