Question

tan square theta minus sin square theta equals to tan square theta sin square theta ​

Answer 1

To Prove :

$\tan^{2}( \theta) - \sin^{2} (\theta) = \tan^{2}( \theta).\sin^{2} (\theta)$

Proof :

• I'm taking RHS to Prove this Identity :

$\longrightarrow\tan^{2}( \theta) \times\sin^{2} (\theta)$

⠀

⠀⠀⠀⠀⋆ tan²( θ ) = sec²( θ ) - 1

⠀

$\longrightarrow(\sec^{2}( \theta) - 1) \times\sin^{2} (\theta)$

⠀

⠀⠀⠀⠀⋆ sec²( θ ) = $\dfrac{1}{\cos^{2}( \theta)}$

⠀

$\longrightarrow\bigg( \dfrac{1}{\cos^{2}( \theta)} - 1\bigg) \times\sin^{2} (\theta)$

$\longrightarrow\bigg( \dfrac{\sin^{2} (\theta)}{\cos^{2}( \theta)} - \sin^{2} (\theta)\bigg)$

⠀

⠀⠀⠀⠀⋆ $\dfrac{\sin^{2} (\theta)}{\cos^{2}( \theta)}$ = tan²( θ )

⠀

$\longrightarrow\tan^{2}( \theta) - \sin^{2} (\theta)$

⠀

$\therefore\boxed{\tan^{2}( \theta) - \sin^{2} (\theta) = \tan^{2}( \theta).\sin^{2} (\theta)}$

Answer 2

To Prove :----

• Tan²a - sin²a = tan²a × sin²a

Formula used :-----

• Tana = sina/cosa
• 1 - cos²a = sin²a
• tana = 1/cota
• sina = 1/coseca

we can solve it by many methods .

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$\pink{\bold{\underline{\underline{<strong>Solution</strong><strong>(</strong><strong>1</strong><strong>)</strong>}}}}$

Lets Try to solve it by LHS First ,,

tan²A - sin²A

putting tan²A = sin²A/cos²A we get,,

→ (sin²A/cos²A) - sin²A

Taking LCM of denominator Now,

→ (sin²A - cos²A×sin²A)/cos²A

Taking sin²A common Now,

→ sin²A[1-cos²A]/cos²A

Putting (1-cos²A = sin²A) now

→ sin²A[sin²A]/cos²A

→ sin²A[sin²A/cos²A]

→ sin²A × tan²A = RHS .

(Proved)

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$\red{\bold{\underline{\underline{Solution(2)}}}}$

Putting tanA = 1/cotA and sinA = 1/cosecA in LHS this time now, we get,

1/cot²A - 1/cosec²A

taking LCM we get,

→ (cosec²A - cot²A)/cot²A×cosec²A

→ Now we know that, cosec²A - cot²A = 1

→ 1/cot²A×cosec²A

→ 1/cot²A × 1/cosec²A

→ tan²A × sin²A = RHS

(Proved)

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(Hope it Helps you)