Prove that- 1-tan2 A1-cot2 A-1-tan Acot A2-tan2A

1+tan^2A1+^2 A=1+tan^2 A1+1tan^2A=(1+tan^2A)tan^2A1+tan^2A=tan ^2A ⇒(1 tan A A)^2=1+sin Acos Acos Asin A ⇒(1 tan A1/tan A)^2=tan A tan^2A=tan^2A ...

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Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

Prove that: ...

Question

Prove that:
$(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{c o t A})^{2} = t a n^{2} A$

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Solution

$\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A} = \frac{1 + {tan}^{2} A}{1 + \frac{1}{{tan}^{2} A}} = \frac{(1 + {tan}^{2} A) {tan}^{2} A}{1 + {tan}^{2} A} = {tan}^{2} A$
$\Rightarrow {(\frac{1 - tan A}{cot A})}^{2} = \frac{1 + \frac{sin A}{cos A}}{\frac{cos A}{sin A}}$
$\Rightarrow {(\frac{1 - tan A}{1 / tan A})}^{2} = tan A - {tan}^{2} A = {tan}^{2} A$

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Question 5 (x)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{1 - c o t A})^{2} = t a n^{2} A$

Q. Prove that

\frac{1 + t a n^{2} A}{1 + c o t^{2} A}

t a n^{2} A .

Q. show that

\frac{1 - {t a n}^{2} A}{{c o t}^{2} A - 1} = {t a n}^{2} A

Q. Prove that

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

Q. Prove

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

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Prove that:(1+tan2A1+cot2A)=(1−tanAcotA)2=tan2A

1+tan2A1+cot2A=1+tan2A1+1tan2A=(1+tan2A)tan2A1+tan2A=tan2A⇒(1−tanAcotA)2=1+sinAcosAcosAsinA⇒(1−tanA1/tanA)2=tanA−tan2A=tan2A

Prove that:
$(\frac{1 + t a n^{2} A}{1 + c o t^{2} A}) = (\frac{1 - t a n A}{c o t A})^{2} = t a n^{2} A$