Question

Prove that: $2 + tan^2A+ \frac{1}{tan^2A}= \frac{1}{sin^2A-sin^4A}$

Answer 1

Answer:

Step-by-step explanation:

Hi,

Consider 2 + tan²A + 1/tan²A

= (1 + tan²A) + ( 1 + 1/tan²A)

= sec²A + (1 + tan²A)/tan²A

= sec²A + sec²A/tan²A

= sec²A(1 + 1/tan²A)

= sec²A( 1 + tan²A)/tan²A

sec⁴A/tan²A

=(1/cos⁴A)/(sin²A/cos²A)

= 1/sin²Acos²A

=1/sin²A(1-sin²A)

=1/sin²A - sin⁴A

= R.H.S

Hence, Proved !

Hope, it helps !

Answer 2

Solution :

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We know that ,

i )cotA = 1/tanA

ii ) 1 + cot²A = cosec²A

iii ) cosecA = 1/sinA

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Now,

LHS = tan²A+ 1/tan²A + 2

= tan²A + cot²A + 2tanAcotA

= ( tanA + cotA )²

= ( 1/cotA + cotA )²

=[ ( 1 + cot²A )/cotA ]²

= [ cosec²A /cot A ]²

= cosec⁴A /cot²A

= ( cosec⁴A )/( cosec² A - 1 )

= ( 1/sin⁴A )/[ ( 1/sin²A ) - 1 ]


= (1/sin⁴A )/[ ( 1 - sin²A )/sin²A ]

= 1/[ sin²A ( 1 - sin²A )]

= 1/( sin²A - sin⁴A )

= RHS

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