Question

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

Answer 1

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

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

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

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

Answer 2

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

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.
• $\frac{7}{25} \: is \: the \: correct \: Answer$