please help I am stuck in this question
Answers
Solution :
Method 1 ( Classical trigonometric approach ) -
• 1/cos theta = sec theta
• 1/sin theta = cosec theta.
• sin theta/cos theta = tan theta
• cos theta/sin theta = cot theta
LHS :
( 1/cos theta - 1/sin theta ) ( 1 + sin theta/cos theta + cos theta/sin theta )
=> [ sec theta - cosec theta ][ 1 + tan theta + cot theta ]
=> sec theta [ 1 + tan theta + cot theta ] - cosec theta [ 1 + tan theta + cot theta ]
=> sec theta + sec theta tan theta + sec theta cot theta - cosec theta - cosec theta tan theta - cosec theta cos theta
=> tan theta [ sec theta - cosec theta ] + cot theta [ sec theta - cosec theta ] + { sec theta - cosec theta ]
Let us simplify sec theta - cosec theta first
sec theta - cosec theta
=> [ 1/cos theta - 1/sin theta ]
=> [ sin theta - cos theta ]/[ sin theta cos theta ]
tan theta [ sec theta - cosec theta ]
=> [ sin theta - cos theta ]/[ cos ² theta ]
cot theta [ sec theta - cosec theta ]
=> [ sin theta - cos theta ]/[ sin ² theta ]
Thus , this simplifies to :
[ sin theta - cos theta ]/[ cos ² theta ] + [ sin theta - cos theta ]/[ sin ² theta ] + [ sin theta - cos theta ]/[ sin theta cos theta ]
Taking [ sin theta - cos theta ] common
=> [ sin theta - cos theta ] { 1/cos² theta + 1/sin² theta + 1/sin theta cos theta }.
Lets simplify { 1/cos² theta + 1/sin² theta + 1/sin theta cos theta } first .
{ 1/cos² theta + 1/sin² theta + 1/sin theta cos theta }
=> [ ( sin² theta + cos² theta )/sin²theta cos theta + 1/sin theta cos theta ]
sin² theta + cos² theta = 1.
=> [ 1/sin² theta cos² theta + 1/sin theta cos theta ]
=> [ sin theta cos theta + 1] /[ sin² theta cos² theta ]
Placing this;
=> [ sin theta - cos theta ] • [ sin theta cos theta + 1] /[ sin² theta cos² theta ]
=> [ ( sin theta - cos theta )( sin theta cos theta + 1 )] / [ sin² theta cos² theta ]
=> [ sin theta ( sin theta cos theta + 1 ) - cos theta ( sin theta cos theta + 1) ]/[ sin² theta cos ² theta ]
=> [ sin² theta cos theta + sin theta - cos² theta sin theta - cos theta ]/[ sin² theta cos ² theta ]
=> sin theta/cos² theta - cos theta / sin² theta
=> ( sin theta / cos theta ) • ( 1/cos theta ) - ( cos theta / sin theta ) • (1/ sin theta ) .
=> tan theta sec theta - cot theta cosec theta
Hence Proved
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Method 2 :
Take sin theta = a and cos theta = b .
This will be a shortcut :)
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