If sin⁴a+sin²a=1 then prove tan⁴a-tan²a=1
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Answered by
12
sin ^4 a = 1 -sin ^2 a
sin^4 a = cos ²a [∵ 1-sin²a = cos²a]
sin^4a/cos^4a = cos²a / cos ^4 a [ dividing each term with cos^4 a]
tan^4 a = 1/cos ²a
tan ^4 a = sec ²a
tan ^4 a = 1+ tan ² a [ ∵ sec²a = 1+tan²a]
tan^4 a - tan² a= 1
hence proved
sin^4 a = cos ²a [∵ 1-sin²a = cos²a]
sin^4a/cos^4a = cos²a / cos ^4 a [ dividing each term with cos^4 a]
tan^4 a = 1/cos ²a
tan ^4 a = sec ²a
tan ^4 a = 1+ tan ² a [ ∵ sec²a = 1+tan²a]
tan^4 a - tan² a= 1
hence proved
Answered by
0
Step-by-step explanation:
Before we begin I say I don't have a sign called theta in my keyboard so i am using A instead of theta. Do use it to ur convenience
sin⁴A+sin²A = 1
sin⁴A+sin²A=sin²A+cos²A
sin⁴A=cos²A
sin⁴/cos⁴A = cos²A/cos⁴A
tan⁴A=1/cos²A
tan⁴A=sec²A
tan⁴A=1+tan²A
tan⁴A-tan²A=1[Proved]
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