Question

prove that (1+1/tan²A) (1+1/cot²A) = 1/sin²A-sin⁴A​

Answer 1

Answer:

LHS = RHS

Step-by-step explanation:

LHS = $[1 +\sf\dfrac{1}{tan^{2}A}] [1 + \sf\dfrac{1}{cot^{2}A}]$

RHS = $\sf\dfrac{1}{sin^{2}A - sin^{4}A}$

Simplifying the LHS:

$\longrightarrow$ $[1 +\sf\dfrac{1}{tan^{2}A}] \ [1 + \sf\dfrac{1}{cot^{2}A}]$

We know that,

$\boxed{\sf{\frac{1}{tan^{2}A} = cot^{2}A}}$

$\boxed{\sf{\frac{1}{cot^{2}A} = tan^{2}A}}$

Substituing these formulas in the equation we get,

$\longrightarrow$  $\sf{[1 + cot^{2}A] [1 + tan^{2}A]}$

We know that

$\large\boxed{\sf{1 + cot^{2}A = cosec^{2}A}}$

$\large\boxed{\sf{1 + tan^{2}A = sec^{2}A}}$

Substituing these formulas in the equation we get

$\longrightarrow$  $\sf{[cosec^{2}A] [1 + sec^{2}A]}$

We know that,

$\boxed{\sf{cosec^{2}A = \dfrac{1}{sin^{2}}}}$

$\boxed{\sf{sec^{2}A = \dfrac{1}{cos^{2}}}}$

Substituing these formulas in the equation we get

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A} \times \dfrac{1}{cos^{2}A}$

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A \times cos^{2}}$

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A \times [1 - sinA^{2}]}$

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A - sin^{4}A}$

∴ LHS = RHS

Hence Proved.

Answer 2

Answer:

LHS = RHS

Step-by-step explanation:

LHS = $[1 +\sf\dfrac{1}{tan^{2}A}] [1 + \sf\dfrac{1}{cot^{2}A}]$

RHS = $\sf\dfrac{1}{sin^{2}A - sin^{4}A}$

Simplifying the LHS:

$\longrightarrow$ $[1 +\sf\dfrac{1}{tan^{2}A}] \ [1 + \sf\dfrac{1}{cot^{2}A}]$

We know that,

$\boxed{\sf{\frac{1}{tan^{2}A} = cot^{2}A}}$

$\boxed{\sf{\frac{1}{cot^{2}A} = tan^{2}A}}$

Substituing these formulas in the equation we get,

$\longrightarrow$  $\sf{[1 + cot^{2}A] [1 + tan^{2}A]}$

We know that

$\large\boxed{\sf{1 + cot^{2}A = cosec^{2}A}}$

$\large\boxed{\sf{1 + tan^{2}A = sec^{2}A}}$

Substituing these formulas in the equation we get

$\longrightarrow$  $\sf{[cosec^{2}A] [1 + sec^{2}A]}$

We know that,

$\boxed{\sf{cosec^{2}A = \dfrac{1}{sin^{2}}}}$

$\boxed{\sf{sec^{2}A = \dfrac{1}{cos^{2}}}}$

Substituing these formulas in the equation we get

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A} \times \dfrac{1}{cos^{2}A}$

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A \times cos^{2}}$

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A \times [1 - sinA^{2}]}$

$\longrightarrow$  $\sf\dfrac{1}{sin^{2}A - sin^{4}A}$

∴ LHS = RHS

Hence Proved.