Question

निम्नलिखित में बहुपद ।
eros de
दो यों वाला राषिक समीकदया है -
Santerbased
एक त्रिभुज के तीनों कोणों का योग होता है
1300670430
आयत का प्रत्येक कोण होता है-
विजत स्थानों की पूर्ति कीजिये​

Answer 1

$Answer:</p><p></p><p>\LARGE{\bf{\underline{\underline{GIVEN:-}}}}GIVEN:−</p><p></p><p>\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}∙  (1+sinA+cosA)2(1+sinA−cosA)2</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 .</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}\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}\end{gathered}sin2θ+cos2θ=11+cot2θ=cosec2θ1+tan2θ=sec2θ</p><p></p><p>T-RATIOS:</p><p></p><p>\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered}∠AsinAcosAtanAcosecAsecAcotA0∘010NotDefined1NotDefined30∘212331232345∘2121122160∘232133223190∘10NotDefined1NotDefined0</p><p>$