please solve this immediately.
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=(sin30°-sin90°+2cos0°)/(tan30°×tan60°)
=(cos(90°-30°)-cos(90°-90°)+2cos0°)/ (cot(90°-30°)×tan60°)
=(cos60°-cos0°+2cos0°)/(cot60°×tan60°)
=(cos60°-cos0°)/1
=cos60°-cos0°
It can be solved further:
=(1/2)-(1)
=-(1/2)
Plz mark it as brainliest...
=(cos(90°-30°)-cos(90°-90°)+2cos0°)/ (cot(90°-30°)×tan60°)
=(cos60°-cos0°+2cos0°)/(cot60°×tan60°)
=(cos60°-cos0°)/1
=cos60°-cos0°
It can be solved further:
=(1/2)-(1)
=-(1/2)
Plz mark it as brainliest...
SOPHIAn:
its not matching with tge given answer
*the
you are great
plz mark it as brainliest...
but its true
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