Question

7. If tan A+
tan A
Solution:
1
2 then show that tan’ A+
= 2
.
tan? A​

Answer 1

Given :

• $\sf tan \: A + \dfrac{1}{tan \: A} = 2$

To prove :

• $\sf tan^2A + \dfrac{1}{tan^2A} = 2$

Proof :

$\implies \sf tan \: A + \dfrac{1}{tan \: A} = 2$

• Squaring both sides

$\implies \sf \left(tan \: A + \dfrac{1}{tan \: A} \right)^2= (2)^2$

• Apply identity
• (a + b)² = a² + b² + 2ab

$\implies \sf (tan \: A)^2 + \left( \dfrac{1}{tan \: A} \right)^2 + 2 \times tan \: A \times \dfrac{1}{ tan \: A}= 4$

$\implies \sf tan^2A + \dfrac{1}{tan^2A} + 2 \times \cancel{ tan \: A }\: \times \dfrac{1}{ \cancel{ tan \: A}} = 4$

$\implies \sf tan^2A + \dfrac{1}{tan^2A} + 2 = 4$

$\implies \sf tan^2A + \dfrac{1}{tan^2A} = 4 - 2$

$\implies \sf tan^2A + \dfrac{1}{tan^2A} = 2$

• $\sf tan^2A + \dfrac{1}{tan^2A} = 2$ ---PROVED

Answer 2

$\huge{\underline{\green{Given:-}}}$

• $tan\:A+\dfrac{1}{tan\:A}=2$

$\huge{\underline{\green{To\:Show:-}}}$

• $tan^2\:A+\dfrac{1}{tan^2\:A}=2$

$\huge{\underline{\green{Identity\:Used:-}}}$

• (a + b)² = a² + b² + 2ab

$\huge{\underline{\green{Solution:-}}}$

$\implies\:tan\:A+\dfrac{1}{tan\:A}=2$

Squaring both sides,

$\implies\:(tan\:A+\dfrac{1}{tan\:A})^2=2^2$

$\implies\sf\:tan^2\:A+\dfrac{1}{tan^2\:A}+2×tan\:A×\dfrac{1}{tan\:A}=4$

$\implies\:tan^2\:A+\dfrac{1}{tan^2\:A}+2=4$

$\implies\:tan^2\:A+\dfrac{1}{tan^2\:A}=4-2$

${\boxed{\implies\:tan^2\:A+\dfrac{1}{tan^2\:A}=2}}$

Hence Proved.

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More Identities:-

• (a - b)² = a² + b² - 2ab

• a² - b² = (a + b)(a - b)

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