hey guys please answer my question and I'll mark u brainliest
Attachments:
Answers
Answered by
1
cosA +sinA = √2cosA
squaring
⇒(cosA + sin A)² = (√2cosA)²
⇒cos²A + sin²A + 2sinAcosA = 2cos²A
⇒1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A
⇒2 - 2cos²A = cos²A + sin²A - 2sinAcosA
⇒2(1 - cos²A)= (cosA - sinA)²
⇒ cosA - sinA = √[2sin²A]
⇒cosA-sinA = √2sinA
hope this helps you
please follow me
squaring
⇒(cosA + sin A)² = (√2cosA)²
⇒cos²A + sin²A + 2sinAcosA = 2cos²A
⇒1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A
⇒2 - 2cos²A = cos²A + sin²A - 2sinAcosA
⇒2(1 - cos²A)= (cosA - sinA)²
⇒ cosA - sinA = √[2sin²A]
⇒cosA-sinA = √2sinA
hope this helps you
please follow me
1402Aryan1402:
instead of converting sin^2@+cos^2@ into 1 . i converted sin^2@ into 1-cos^2@ and cos^2@into 1-sin^2@
then i rearranged the equation
to get the desired result
to say u can solve this in a more simpler and short way.
@prasanthrai its correct! and to tell these type of questions are important once
it should be √2sinA instead of only √2 in 'To show' side
Answered by
1
HERE IS YOUR ANSWER NO NEED OF ANY SQUARING,SIMPLE ,
HOPE IT HELPS YOU.
HOPE IT HELPS YOU.
Attachments:
Similar questions