IT FULL SOLUTION STEP BY STEP.
Attachments:
Answers
Answered by
1
Heya!!✋
An outstanding question you have asked from trigonometry. Here is your answer. Please, consider theta as A
================================
{(sinA + cosA)\(sinA - cosA)}+{(sinA-cosA)\(sinA+cosA)}
=> {(sinA + cosA)²+(sinA-cosA)²}\(sin²A-cos²A)
=>( sin²A + 2sinAcosA+cos²A+sin²A-2sinAcosA+cos²A)\sin²A-cos²A
=> 2sin²A+2cos²A\ sin²A-cos²A
=> 2(sin²A+cos²A)\ sin²A-cos²A
=> 2×1\ sin²A-cos²A
=> 2\ sin²A-cos²A
=> (2\sin²A)-(2cos²A)
=>{ (2\cos²A)\(sin²A\cos²A)}-{(2cos²A)\(cos²A\cos²A)}
=> (2sec²A\tan²A)-2sec²A
=> 2sec²A\(tan²A-1)
[Proved]
=================================
Friend,I hope you can solve such type of problem from now ^_^
An outstanding question you have asked from trigonometry. Here is your answer. Please, consider theta as A
================================
{(sinA + cosA)\(sinA - cosA)}+{(sinA-cosA)\(sinA+cosA)}
=> {(sinA + cosA)²+(sinA-cosA)²}\(sin²A-cos²A)
=>( sin²A + 2sinAcosA+cos²A+sin²A-2sinAcosA+cos²A)\sin²A-cos²A
=> 2sin²A+2cos²A\ sin²A-cos²A
=> 2(sin²A+cos²A)\ sin²A-cos²A
=> 2×1\ sin²A-cos²A
=> 2\ sin²A-cos²A
=> (2\sin²A)-(2cos²A)
=>{ (2\cos²A)\(sin²A\cos²A)}-{(2cos²A)\(cos²A\cos²A)}
=> (2sec²A\tan²A)-2sec²A
=> 2sec²A\(tan²A-1)
[Proved]
=================================
Friend,I hope you can solve such type of problem from now ^_^
jmsbrothers:
Can you mark it as brainliest?
:)
why have you divided it bu cos^2
in last 3rd step
I divided cos²A so that tan²A can come in the equation ^_^
do u have any question? Ask me..np
2\sin²-cos² ∆ (2\cos²)\{(sin²-cos²)\cos²} ∆ 2sec²\{(sin²\cos²)-(cos²\cos²)} ∆ 2sec²\(tan²-1)
hope u will understand better now
Similar questions