Question

∙  (1+sinA+cosA)2(1+sinA−cosA)2​</p><p>

\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}SOLUTION:−​​

LHS:

\sf \to \dfrac{(1+sinA-cosA)^2}{(1+sinA+cosA)^2}→(1+sinA+cosA)2(1+sinA−cosA)2​

Expand the fractions using .

\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)​

Rearrange the terms.

\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)​

We know that cos²A+sin²A=1.

\sf \to \dfrac{1-2sincos-2cos}{2sin+1}→2sin+11−2sincos−2cos​

Now here, take -2cos common from the numerator and +2cos common from the denominator.

\sf \to \dfrac{1-2cos(sin+2)}{2sin+1}→2sin+11−2cos(sin+2)​

Now, rearrange the terms, add 1 and 1 and take 2 common.

\to\sf\dfrac{1+1+2sin-2cos}{sin+1}→sin+11+1+2sin−2cos​

\to\sf\dfrac{2+2sin-2cos}{sin+1}→sin+12+2sin−2cos​

Take 2 common.

\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)​

Take (1+sin) common.

\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)​

\to \sf{\red{\dfrac{1-cosA}{1+cosA} }}→1+cosA1−cosA​

LHS=RHS.

HENCE PROVED!

FUNDAMENTAL TRIGONOMETRIC RATIOS:

\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​​​​​

T-RATIOS:

\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} < /p > < p > \end{gathered}∠AsinAcosAtanAcosecAsecAcotA​0∘010NotDefined1NotDefined​30∘21​23​​3​1​23​2​3​​45∘2​1​2​1​12​2​1​60∘23​​21​3​3​2​23​1​​90∘10NotDefined1NotDefined0​​



free points
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Answer 1

Answer:

Consider the LHS: (1-sinA+cosA)2 = [(1-sinA) + cosA]2 = (1-sinA)2 + cos2A + 2(1-sinA)cosA = 1 + sin2A − 2sinA + cos2A + 2(1-sinA)cosA = 1 + (sin2A + cos2A) − 2sinA + 2(1-sinA)cosA = 1 + 1 − 2sinA + 2(1-sinA)cosA [Since, sin2A + cos2A =1] = 2 − 2sinA + 2(1-sinA)cosA = 2(1 − sinA) + 2(1-sinA)cosA = 2(1 − sinA)(1 + cosA) = RHS