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Class 10 exercise 8.4 ques 5 |
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Answer» (i) (cosec θ – cot θ)2\xa0= (1-cos θ)/(1+cos θ)To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)L.H.S. = (cosec θ – cot θ)2The above equation is in the form of (a-b)2, and expand itSince (a-b)2\xa0= a2\xa0+ b2\xa0– 2abHere a = cosec θ and b = cot θ= (cosec2θ +\xa0cot2θ – 2cosec θ cot θ)Now, apply the corresponding inverse functions and equivalent ratios to simplify= (1/sin2θ\xa0+ cos2θ/sin2θ – 2cos θ/sin2θ)= (1 + cos2θ – 2cos θ)/(1 – cos2θ)= (1-cos θ)2/(1 – cosθ)(1+cos θ)= (1-cos θ)/(1+cos θ) = R.H.S.Therefore, (cosec θ – cot θ)2\xa0= (1-cos θ)/(1+cos θ)Hence proved.(ii) (cos A/(1+sin A))\xa0+ ((1+sin A)/cos A) = 2 sec ANow, take the L.H.S of the given equation.L.H.S. = (cos A/(1+sin A))\xa0+ ((1+sin A)/cos A) = [cos2A\xa0+ (1+sin A)2]/(1+sin A)cos A = (cos2A + sin2A + 1\xa0+ 2sin A)/(1+sin A) cos ASince cos2A + sin2A = 1, we can write it as = (1\xa0+ 1 + 2sin A)/(1+sin A) cos A = (2+ 2sin A)/(1+sin A)cos A = 2(1+sin A)/(1+sin A)cos A = 2/cos A = 2 sec A = R.H.S.L.H.S. = R.H.S.(cos A/(1+sin A))\xa0+ ((1+sin A)/cos A) = 2 sec AHence proved.\xa0For more click on the given link:NCERT Solutions for Class 10 Maths Exercise 8.4 ... Yes We have to solve all parts of question 5 |
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