Math, asked by harsh2576, 5 months ago

please help I am stuck in this question​

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Answered by Saby123
10

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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