Question

probe that (1+sin-cos)2/(1+sin+cos)2=1-cos/1+cos
​

Answer 1

Answer:

$1+sinA−cosA)2=1+cos1−cos</p><p></p><p>\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}SOLUTION:−</p><p></p><p>LHS:</p><p></p><p>\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→(1+sinA+cosA)2(1+sinA−cosA)2</p><p></p><p>Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.</p><p></p><p>\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→(cos2+2sincos+sin2+2cos+2sin+1)(cos2−2sincos+sin2−2cos+2sin+1)</p><p></p><p>Rearrange the terms.</p><p></p><p>\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→(cos2+sin2+2sincos+2cos+2sin+1)(cos2+sin2−2sincos−2cos+2sin+1)</p><p></p><p>We know that cos²A+sin²A=1.</p><p></p><p>\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→2sin+11−2sincos−2cos</p><p></p><p>Now here, take -2cos common from the numerator and +2cos common from the denominator.</p><p></p><p>\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→2sin+11−2cos(sin+2)</p><p></p><p>Now, rearrange the terms, add 1 and 1 and take 2 common.</p><p></p><p>\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→sin+11+1+2sin−2cos</p><p></p><p>\to\sf\dfrac{2+2sin-2cos}{sin+1}→sin+12+2sin−2cos</p><p></p><p>Take 2 common.</p><p></p><p>$

Step-by-step explanation:

mark as brainlist answer