श्री रामदास को श्रीलंका में क्या क्या पढ़ने और पढ़ने का अवसर प्राप्त हुए
Answers
Answer:
Solution :
\displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{cot^3x-tan\ x}{cos\left( \frac{\pi}{4} + x \right)}
x→
4
π
lim
cos(
4
π
+x)
cot
3
x−tan x
On plugging the values directly into the eq. , it leads to indeterminant form \dfrac{0}{0}
0
0
\bullet\ \; \sf \orange{cot\ x = \dfrac{1}{tan\ x}}∙ cot x=
tan x
1
\bullet\ \; \sf \green{cos(A+B)=cos\ A\ .cos\ B-sin\ A\ .sin\ B}∙ cos(A+B)=cos A .cos B−sin A .sin B
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ \frac{1}{tan^3x} -tan\ x}{cos\ \frac{\pi}{4}.cos\ x- sin\ \frac{\pi}{4}.sin\ x}⟶
x→
4
π
lim
cos
4
π
.cos x−sin
4
π
.sin x
tan
3
x
1
−tan x
\bullet\ \; \sf \blue{cos\ \dfrac{\pi}{4}=sin\ \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}}∙ cos
4
π
=sin
4
π
=
2
1
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ \frac{1-tan^4x}{tan^3x}}{\frac{1}{\sqrt{2}}\left(cos\ x-sin\ x\right)}⟶
x→
4
π
lim
2
1
(cos x−sin x)
tan
3
x
1−tan
4
x
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ 1-tan^4x}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}}{tan^3x}⟶
x→
4
π
lim
(cos x−sin x)
1−tan
4
x
×
tan
3
x
2
\bullet\ \; \sf \orange{1-A^4=1^2-(A^2)^2=(1-A^2)(1+A^2)}∙ 1−A
4
=1
2
−(A
2
)
2
=(1−A
2
)(1+A
2
)
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ 1-tan^2x}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(1+tan^2x)}{tan^3x}⟶
x→
4
π
lim
(cos x−sin x)
1−tan
2
x
×
tan
3
x
2
(1+tan
2
x)
\bullet\ \; \sf \green{sec^2x=1+tan^2x}∙ sec
2
x=1+tan
2
x
\bullet\ \; \sf \blue{tan\ x=\dfrac{sin\ x}{cos\ x}}∙ tan x=
cos x
sin x
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ 1-\frac{sin^2x}{cos^2x}}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(sec^2x)}{tan^3x}⟶
x→
4
π
lim
(cos x−sin x)
1−
cos
2
x
sin
2
x
×
tan
3
x
2
(sec
2
x)
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ cos^2x-sin^2x}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(sec^2x)}{cos^2x.tan^3x}⟶
x→
4
π
lim
(cos x−sin x)
cos
2
x−sin
2
x
×
cos
2
x.tan
3
x
2
(sec
2
x)
\begin{gathered}\bullet\ \; \sf \orange{sec^2x=\dfrac{1}{cos^2x}}\\\\\bullet\ \; \sf \green{cos^2x=\dfrac{1}{sec^2x}}\end{gathered}
∙ sec
2
x=
cos
2
x
1
∙ cos
2
x=
sec
2
x
1
\begin{gathered}\\\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ (cos\ x-sin\ x)(cos\ x+sin\ x)}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(sec^4x)}{tan^3x}\\\end{gathered}
⟶
x→
4
π
lim
(cos x−sin x)
(cos x−sin x)(cos x+sin x)
×
tan
3
x
2
(sec
4
x)
\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} (cos\ x+sin\ x )\left(\dfrac{\sqrt{2}(sec^4x)}{tan^3x}\right)⟶
x→
4
π
lim
(cos x+sin x)(
tan
3
x
2
(sec
4
x)
)
Sub. the limit value ,
\longrightarrow \sf \left( cos\ \dfrac{\pi}{4}+sin\ \dfrac{\pi}{4} \right) \left(\dfrac{\sqrt{2}(sec^4 \frac{\pi}{4} )}{tan^3 \frac{\pi}{4} }\right)⟶(cos
4
π
+sin
4
π
)(
tan
3
4
π
2
(sec
4
4
π
)
)
\bullet\ \; \sf \blue{cos\ \dfrac{\pi}{4}=sin\ \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}}∙ cos
4
π
=sin
4
π
=
2
1
\bullet\ \; \sf \orange{sec\ \dfrac{\pi}{4}= \sqrt{2}}∙ sec
4
π
=
2
\bullet\ \; \sf \green{tan\ \dfrac{\pi}{4}=1}∙ tan
4
π
=1
\longrightarrow \sf \left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right) \left(\dfrac{\sqrt{2}(\sqrt{2})^4}{(1)^3 }\right)⟶(
2
1
+
2
1
)(
(1)
3
2
(
2
)
4
)
\longrightarrow \sf \dfrac{2}{\sqrt{2}} \times 4\sqrt{2}⟶
2
2
×4
2
\longrightarrow\ \; \sf \pink{8}⟶ 8
★ ═════════════════════ ★
\bullet\ \; \displaystyle \bf \purple{\lim_{x \to \frac{\pi}{4}} \dfrac{cot^3x-tan\ x}{cos\left( \frac{\pi}{4} + x \right)} =8}∙
x→
4
π
lim
cos(
4
π
+x)
cot
3
x−tan x
=8