tan square theta minus sin square theta equals to tan square theta sin square theta
Answers
To Prove :
Proof :
• I'm taking RHS to Prove this Identity :
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⠀⠀⠀⠀⋆ tan²( θ ) = sec²( θ ) - 1
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⠀⠀⠀⠀⋆ sec²( θ ) =
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⠀⠀⠀⠀⋆ = tan²( θ )
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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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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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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)