Math, asked by yash191919, 4 months ago

prove that (1 + cot A + tan A)( Sin A - Cos A)/ sec^3 A -cosec^3 A =sin^2A cos^2A​​

Answers

Answered by Anonymous
1

 \sf \mathfrak{\huge \underline{To~Prove:}}

 \sf \frac {(1 + cot A + tan A)( Sin A - Cos A)} {sec^3 A -cosec^3 A} =sin^2A cos^2A

 \mathfrak{\huge \underline{Solution:}}

 \sf \frac {(1 + cot A + tan A)( Sin A - Cos A)} {sec^3 A -cosec^3 A} =sin^2A cos^2A \\\\\sf \longrightarrow\frac {(1+\frac{cosA} {sinA}+\frac{sinA}{cosA}) (SinA-CosA)}{\frac{1}{cos^3A}-\frac{1}{sin^3A}} \\\\\sf\longrightarrow \frac{(\frac{sinAcosA+cos^2A+sin^2A} {sinAcosA})(sinA-cosA)}{\frac{sin^3A-cos^3A}{cos^3Asin^3A}} \\\\\sf\longrightarrow \frac{(sinAcosA+1) (sinA-cosA)}{sinAcosA} \times \frac{cos^3Asin^3A}{sin^3A-cos^3A} \\\\\sf\longrightarrow \frac{(sinAcosA+1) (sinA-cosA)}{sinAcosA} \times \frac{cosAsinA(cos^2Asin^2A)}{sin^3A-cos^3A} \\\\\sf \longrightarrow\frac{(sinAcosA+1)(sinA-cosA)}{\cancel{sinAcosA}} \times \frac{\cancel{cosAsinA}(cos^2Asin^2A)}{sin^3A-cos^3A} \\\\\sf \longrightarrow \frac{(sinAcosA+1)(sinA-cosA)(cos^2Asin^2A)}{sin^3A-cos^3A}\\\\\sf By~identity \\\\\sf \boxed{a^3-b^3=(a-b)(a^2+b^2+ab)} \\\\\sf sin^3A-cos^3A= (sinA-cosA) (sin^2A+cos^2A+sin^3Acos^3A) \\\\\sf so, further~ simplification ~is~as~follows: \\\\\sf \longrightarrow \frac{(sinAcosA+1)(sinA-cosA)(cos^2Asin^2A)}{(sinA-cosA) (sin^2A+cos^2A+sin^3Acos^3A)}\\\\\sf \longrightarrow \frac{(sinAcosA+1)\cancel {(sinA-cosA)} (cos^2Asin^2A)}{\cancel {(sinA-cosA)} (sin^2A+cos^2A+sinAcosA)}\\\\\sf \longrightarrow \frac{(sinAcosA+1)(cos^2Asin^2A)}{1+sinAcosA} \\\\\sf \longrightarrow \frac{(\cancel{sinAcosA+1})(cos^2Asin^2A)}{\cancel{1+sinAcosA}} \\\\\sf \longrightarrow \huge\boxed{cos^2Asin^2A}

Answered by ItZzPriyanka
8

 \large  { \underline{ \bf \orange{Question:- }}}

prove that (1 + cot A + tan A)( Sin A - Cos A)/ sec^3 A -cosec^3 A =sin^2A cos^2A

 \large  { \underline{ \bf \pink{Solution:- }}}

 \large  { \underline{ \bf {Given:- }}}

3cotA = 4 \\ sinA =  \frac{bc}{ac} \\ tanA =  \frac{ab}{ac}  \\ cotA =  \frac{ab}{bc}  =  \frac{4k}{3k}  \\AB=4k \\BC=3k</p><p>

 \large  { \underline{ \bf \green{By \: pythagoras \: Theorem \: we \: get,:- }}}

AC2=AB2+BC2 \\AC2=42+32 \\AC2=16+9 \\ AC2=25 \\ AC2=25 \\AC=5k</p><p>

 \large  { \underline{ \bf \blue{L.H.S:- }}}

  = \frac{1 -tan² A}{1  + tan² A}\\  =    \frac{1 -  (\frac{3}{4})² }{1 +  (\frac{3}{4})² }\\  = \frac{1 -  \frac{9}{16} }{1 +  \frac{9}{16} }\\  =  \frac{16 - 9}{16 + 9} \\  =  \frac{7}{25}

 \large  { \underline{ \bf \pink{R.H.S:- }}}

 = cos²A - sin²A \\  =  ( \: \frac{4}{5} \: ) ² \:  \:  - ( \:  \frac{3}{5}  \: )² \\  =  \frac{16}{25}  -  \frac{9}{25}  \\  =  \frac{7}{25}

 \large  { \underline{ \bf \red{Therefore:- }}}

  • L.H.S=R.H.S proved.
  • \frac{7}{25} \:  is  \: the  \: correct  \: Answer
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