Question

The dream of the 14 year old girl in Hydrabad

Answer 1

Answer:

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}} </p><p>GIVEN:−</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\sf \bullet \ \ \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}∙ </p><p>(1+sinA+cosA) </p><p>2</p><p> </p><p>(1+sinA−cosA) </p><p>2</p><p> </p><p> </p><p> </p><p></p><p>\LARGE{\bf{\underline{\underline{SOLUTION:-}}}} </p><p>SOLUTION:−</p><p> </p><p> </p><p> </p><p> </p><p></p><p>LHS:</p><p></p><p>\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→ </p><p>(1+sinA+cosA) </p><p>2</p><p> </p><p>(1+sinA−cosA) </p><p>2</p><p> </p><p> </p><p> </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)}→ </p><p>(cos </p><p>2</p><p> +2sincos+sin </p><p>2</p><p> +2cos+2sin+1)</p><p>(cos </p><p>2</p><p> −2sincos+sin </p><p>2</p><p> −2cos+2sin+1)</p><p> </p><p> </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)}→ </p><p>(cos </p><p>2</p><p> +sin </p><p>2</p><p> +2sincos+2cos+2sin+1)</p><p>(cos </p><p>2</p><p> +sin </p><p>2</p><p> −2sincos−2cos+2sin+1)</p><p> </p><p> </p><p></p><p>We know that cos²A+sin²A=1.</p><p></p><p>\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→ </p><p>2sin+1</p><p>1−2sincos−2cos</p><p> </p><p> </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}→ </p><p>2sin+1</p><p>1−2cos(sin+2)</p><p> </p><p> </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}→ </p><p>sin+1</p><p>1+1+2sin−2cos</p><p> </p><p> </p><p></p><p>\to\sf\dfrac{2+2sin-2cos}{sin+1}→ </p><p>sin+1</p><p>2+2sin−2cos</p><p> </p><p> </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 ) }→ </p><p>2(1+sin)+2cos(sin+1)</p><p>2(1+sin)−2cos(sin+1)</p><p> </p><p> </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 )}→ </p><p></p><p>2 </p><p>(1+sin)</p><p> </p><p> (1+cos)</p><p></p><p>2 </p><p>(1+sin)</p><p> </p><p> (1−cos)</p><p> </p><p> </p><p></p><p>\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→ </p><p>1+cosA</p><p>1−cosA</p><p> </p><p> </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} </p><p>sin </p><p>2</p><p> θ+cos </p><p>2</p><p> θ=1</p><p>1+cot </p><p>2</p><p> θ=cosec </p><p>2</p><p> θ</p><p>1+tan </p><p>2</p><p> θ=sec </p><p>2</p><p> θ</p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </p><p> </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} </p><p>∠A</p><p>sinA</p><p>cosA</p><p>tanA</p><p>cosecA</p><p>secA</p><p>cotA</p><p> </p><p> </p><p>0 </p><p>∘</p><p> </p><p>0</p><p>1</p><p>0</p><p>NotDefined</p><p>1</p><p>NotDefined</p><p> </p><p> </p><p>30 </p><p>∘</p><p> </p><p>2</p><p>1</p><p> </p><p> </p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> </p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p>2</p><p>3</p><p> </p><p> </p><p>2</p><p> </p><p> </p><p>3</p><p> </p><p> </p><p> </p><p> </p><p>45 </p><p>∘</p><p> </p><p>2</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p>2</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p>1</p><p>2</p><p> </p><p> </p><p>2</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p>60 </p><p>∘</p><p> </p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> </p><p>2</p><p>1</p><p> </p><p> </p><p>3</p><p> </p><p> </p><p>3</p><p> </p><p> </p><p>2</p><p> </p><p> </p><p>2</p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> </p><p> </p><p> </p><p>90 </p><p>∘</p><p> </p><p>1</p><p>0</p><p>NotDefined</p><p>1</p><p>NotDefined</p><p>0$