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Cosec ²A - tan²(90-A)= sin²A +sin(90-A)
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)
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SOLUTION:
cosec²A - tan²(90°-A)= sin²A +sin(90°-A)
LHS=Cosec ²A - tan²(90°-A)
= 1/sin²A - sin²(90°-A)/ cos²(90°-A)
[ cosecA= 1/sinA, tanA= sinA/cosA]
= 1/sin²A - sin²(90°-A)/ sin²A
[ cos(90°-A)= sinA]
= 1/sin²A - cos²A/sin²A
[ sin(90°-A)= cosA]
= 1-cos²A/sin²A
= sin²A/sin²A
[ 1-cos²A= sin²A]
= 1
= sin²A + cos²A [1= sin²A + cos²A ]
= sin²A + sin²(90°-A)
= RHS
HOPE THIS WILL HELP YOU.....
cosec²A - tan²(90°-A)= sin²A +sin(90°-A)
LHS=Cosec ²A - tan²(90°-A)
= 1/sin²A - sin²(90°-A)/ cos²(90°-A)
[ cosecA= 1/sinA, tanA= sinA/cosA]
= 1/sin²A - sin²(90°-A)/ sin²A
[ cos(90°-A)= sinA]
= 1/sin²A - cos²A/sin²A
[ sin(90°-A)= cosA]
= 1-cos²A/sin²A
= sin²A/sin²A
[ 1-cos²A= sin²A]
= 1
= sin²A + cos²A [1= sin²A + cos²A ]
= sin²A + sin²(90°-A)
= RHS
HOPE THIS WILL HELP YOU.....
Jahnavi444:
How can u add sin²(90-A) at the end when there is only sin(90-A) in the question.
mistake is in the question...
why wrong question is posted??
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