Question

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

Answer:

$1+sinA−cosA)2=1+cos1−cos</p><p></p><p>\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}SOLUTION:−</p><p></p><p>LHS:</p><p></p><p>\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→(1+sinA+cosA)2(1+sinA−cosA)2</p><p></p><p>Expand the fractions using (a+b+c)²=a²+b²+c²+2ab+2bc+2ca.</p><p></p><p>\sf \to \dfrac{(cos^2-2sincos+sin^2-2cos+2sin+1)}{(cos^2+2sincos+sin^2+2cos+2sin+1)}→(cos2+2sincos+sin2+2cos+2sin+1)(cos2−2sincos+sin2−2cos+2sin+1)</p><p></p><p>Rearrange the terms.</p><p></p><p>\sf \to \dfrac{(cos^2+sin^2-2sincos-2cos+2sin+1)}{(cos^2+sin^2+2sincos+2cos+2sin+1)}→(cos2+sin2+2sincos+2cos+2sin+1)(cos2+sin2−2sincos−2cos+2sin+1)</p><p></p><p>We know that cos²A+sin²A=1.</p><p></p><p>\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→2sin+11−2sincos−2cos</p><p></p><p>Now here, take -2cos common from the numerator and +2cos common from the denominator.</p><p></p><p>\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→2sin+11−2cos(sin+2)</p><p></p><p>Now, rearrange the terms, add 1 and 1 and take 2 common.</p><p></p><p>\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→sin+11+1+2sin−2cos</p><p></p><p>\to\sf\dfrac{2+2sin-2cos}{sin+1}→sin+12+2sin−2cos</p><p></p><p>Take 2 common.</p><p></p><p>\to \sf \dfrac{ 2(1+sin) -2cos(sin+1) }{ 2(1+sin) + 2cos(sin +1 ) }→2(1+sin)+2cos(sin+1)2(1+sin)−2cos(sin+1)</p><p></p><p>Take (1+sin) common.</p><p></p><p>\to \sf \dfrac{ \not{2}\cancel{(1+sin)}(1 - cos) }{\not{2}\cancel{(1+sin )}(1 + cos )}→2(1+sin)(1+cos)2(1+sin)(1−cos)</p><p></p><p>\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→1+cosA1−cosA</p><p></p><p>LHS=RHS.</p><p></p><p>HENCE PROVED!</p><p></p><p>FUNDAMENTAL TRIGONOMETRIC RATIOS:</p><p></p><p>\begin{gathered}\begin{gathered}\boxed{\substack{\displaystyle \sf sin^2 \theta+cos^2 \theta = 1 \\\\ \displaystyle \sf 1+cot^2 \theta=cosec^2 \theta \\\\ \displaystyle \sf 1+tan^2 \theta=sec^2 \theta}}\end{gathered}\end{gathered}sin2θ+cos2θ=11+cot2θ=cosec2θ1+tan2θ=sec2θ</p><p></p><p></p><p>$

Step-by-step explanation:

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