\(\mathrm{LHS}  =\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A} \quad\left[\because \cot A=\frac{\cos A}{\sin A}\right]\)
\(=\frac{\cos A-\cos A \sin A}{\cos A+\cos A \sin A}=\frac{\cos A(1-\sin A)}{\cos A(1+\sin A)} \)
\( =\frac{(1-\sin A)}{(1+\sin A)} \frac{(1+\sin A)}{(1+\sin A)}=\frac{1-\sin ^2 A}{(1+\sin A)^2} \)
\( =\frac{\cos ^2 A}{(1+\sin A)^2}=\text { RHS } \quad \text { Hence proved. }\)

Let AD be the altitude from A on BC. Then, using pythagoras theorem in MDB and MDC, then find BD and DC.14 cm

In \(\Delta ABC\),
tan 60° =\(\frac{AB}{BC}\)
AB=\(\sqrt{3}\)BC...(i)
In \(\Delta ABD\),
tan 30° =\(\frac{AB}{BC+40}\)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { \sqrt { 3 } BC }{ BC+40 } \)
BC + 40 = 3BC
40 = 2BC
BC = 20m
AB=20\(\sqrt{3}\)m
Height of tower =20\(\sqrt{3}\)m

Given, in \(\triangle ABC\), \(DE\parallel AC\)   [given]
So,  \(\frac { BE }{ EC } =\frac { BD }{ DA } \)  ... (i)
[ by basic proportionality theorem]
Also, \(DC\parallel AP\)    [given]
So, \(\frac { BC }{ CP } =\frac { BD }{ DA } \)  ... (ii)
[ by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac { BE }{ EC } =\frac { BC }{ CP } \)

3 cosec A = 4 cot A 

In \(\triangle\)CAB,
A=B   (Given)
\(\therefore\) AC=CB  (By isosceles triangle property)
But, AD=BE     (Given)...(i)
\(\Rightarrow\)AC-AD=CB-BE
\(\therefore\) CD=CE    ...(ii)
Dividing Equation (ii) by (i),
\(\frac { CD }{ AD } =\frac { CE }{ BE } \)
By converse of BPT,
DE || AB.

\({ cosec }^{ 2 }\left( { 90 }^{ ° }-\theta \right) =\sec ^{ 2 }{ \theta } \)
\(\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \theta =1 } \)
\(\cos ^{ 2 }{ { 40 }^{ ° } } +\cos ^{ 2 }{ { 50 }^{ ° } } =\cos ^{ 2 }{ \left( { 90 }^{ ° }-{ 50 }^{ ° } \right) } +\cos ^{ 2 }{ { 50 }^{ ° } } \)
\(\sin ^{ 2 }{ { 50 }^{ ° } } +\cos ^{ 2 }{ { 50 }^{ ° } } \varpi 1\)
\(\tan ^{ 2 }{ { 30 }^{ ° } } ={ \left( \frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }=\frac { 1 }{ 3 } \)
\(\sec ^{ 2 }{ { 52 }^{ ° } } \sin ^{ 2 }{ { 38 }^{ ° } } =\sec ^{ 2 }{ { 52 }^{ ° } } \sin ^{ 2 }{ \left( { 90 }^{ ° }-{ 52 }^{ ° } \right) } \)
\(\sec ^{ 2 }{ { 52 }^{ ° } } \cos ^{ 2 }{ { 52 }^{ ° } } =1\) 
and \({ cosec }^{ 2 }{ 70 }^{ ° }-\tan ^{ 2 }{ { 20 }^{ ° } } ={ cosec }^{ 2 }\left( { 90 }^{ ° }-{ 20 }^{ ° } \right) -\tan ^{ 2 }{ { 20 }^{ ° } } \)
\(=\sec ^{ 2 }{ { 20 }^{ ° } } -\tan ^{ 2 }{ { 20 }^{ ° } } =1\)
Given expression = \(\frac { 1 }{ 4 } -\frac { 2\times \frac { 1 }{ 3 } \times 1 }{ 3\left( 1 \right) } \)
\(=\frac { 1 }{ 4 } -\frac { 2 }{ 9 } =\frac { 9-8 }{ 36 } =\frac { 1 }{ 36 } \)

In \(\triangle\)ABC, B=90⁰ and AD and CE are two medians
AC² = AB² + BC² = (5)² = 25    (By Pythagoras theorem)  ...(i)
In \(\triangle\)ABD, AD² = AB² + BD²
\(\Rightarrow { \left( \frac { 3\sqrt { 5 } }{ 2 } \right) }^{ 2 }=AB^{ 2 }+\frac { BC^{ 2 } }{ 4 } \)
\(\Rightarrow \frac { 45 }{ 4 } =AB^{ 2 }+\frac { BC^{ 2 } }{ 4 } ...(ii)\)

In \(\triangle\)EBC,   \(CE^{ 2 }=BC^{ 2 }+\frac { AB^{ 2 } }{ 4 } ...(iii)\)
Subtracting equation (ii) from equation (i),
\(\frac { 3BC^{ 2 } }{ 4 } =25-\frac { 45 }{ 4 } =\frac { 55 }{ 4 } \)
\(\Rightarrow BC^{ 2 }=\frac { 55 }{ 3 } ..(iv)\)
From eqn. (ii), 
\(AB^{ 2 }+\frac { 55 }{ 12 } =\frac { 45 }{ 4 } \)
\(\Rightarrow AB^{ 2 }=\frac { 45 }{ 4 } -\frac { 55 }{ 12 } =\frac { 20 }{ 3 } \)
From eqn. (iii),  \({ CE }^{ 2 }=\frac { 55 }{ 3 } +\frac { 20 }{ 3\times 4 } \)
\(=\frac { 240 }{ 12 } =20\)
\(\therefore CE=\sqrt { 20 } =2\sqrt { 5 } cm\)

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

If Assertion is incorrect but Reason is correct.

(i) (c): Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post.
From the figure, DE is the shadow of the girl. Let DE be x metres.
Now, her distance from the base of the lamp = BD = 1.2 m x 4 = 4.8 m.
(ii) (a): In \(\triangle\) ABE and \(\triangle\)CDE, \(\angle\)B = \(\angle\)D ( Each is of 90^o)
and \(\angle\)E = \(\angle\)E (same angle)
So, \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\) (AA similarity criterion)
(iii) (d): \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\)
\(\Rightarrow \frac{\mathrm{BE}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{CD}}\)
\(\Rightarrow \frac{4.8+x}{x}=\frac{3.6}{0.9} \quad \Rightarrow 4.8+x=4 x\)
\(\Rightarrow 3 x=4.8 \ \Rightarrow \ x=1.6\)
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.
(iv) (b): Since ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides,
Ratio of areas of similar triangles = \(\sqrt{9}\) : \(\sqrt{16}\)
= 3 : 4
(v) (b): \(\frac{A E}{C E}=\frac{B E}{D E}=\frac{4.8+1.6}{1.6}=\frac{6.4}{1.6}=4\)
\(\Rightarrow\) AE = 4 CE
\(\Rightarrow\) AC + CE = 4 CE
\(\Rightarrow A C=3 C E \Rightarrow \frac{A C}{C E}=\frac{3}{1}\)

We have, AB = BC = 6\(\sqrt{2}\) m and AC=12m .
(i) (d):\(\because\)  Dis mid point of AC.
\(\therefore\) AD=DC=6m
Now, AB² = BD² + AD² (\(\therefore\) \(\Delta\)ABD is a right triangle)
\(\Rightarrow B D^{2}=(6 \sqrt{2})^{2}-6^{2}=72-36=36 \)
\(\Rightarrow B D=6 \mathrm{~m}\)
(ii) (c) : \(\operatorname{In} \Delta A B D, \sin A=\frac{B D}{A B}=\frac{6}{6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \sin A=\sin 45^{\circ} \Rightarrow \angle A=45^{\circ}\)
(iii) (c) : \(\operatorname{In} \Delta B D C, \tan C=\frac{B D}{D C}=\frac{6}{6}\)
\(\Rightarrow \tan C=1=\tan 45^{\circ} \Rightarrow \angle C=45^{\circ}\)
(iv) (d) : \(\sin A=\frac{1}{\sqrt{2}}, \cos C=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\therefore \quad \sin A+\cos C=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
(v) (c): \((v) \quad(c): \tan C=1, \tan A=\tan 45^{\circ}=1\)
\(\Rightarrow \tan ^{2} C+\tan ^{2} A=1+1=2\)
 

\(\because \Delta\)PQR is a right angled triangle.
\(\therefore\) PR² + RQ² = PQ²
\(\Rightarrow P R^{2}=(13)^{2}-(12)^{2}=25 \Rightarrow P R=5 \mathrm{~cm}\)
(i) (c) : \(\cos \theta=\frac{Q R}{P Q}=\frac{12}{13} \)
(ii) (c) : \(\sec \theta=\frac{1}{\cos \theta}=\frac{13}{12} \)
(iii) (c) : \(\tan \theta=\frac{P R}{R Q}=\frac{5}{12}\)
\(\therefore \frac{\tan \theta}{1+\tan ^{2} \theta}=\frac{\frac{5}{12}}{1+\frac{25}{144}}=\frac{\frac{5}{12}}{\frac{169}{144}}=\frac{60}{169}\)
(iv) (a): \(\cot \theta=\frac{1}{\tan \theta}=\frac{12}{5}\)    [Using (1)]
\(\operatorname{cosec} \theta=\frac{P Q}{P R}=\frac{13}{5} \)
\(\therefore \quad \cot ^{2} \theta-\operatorname{cosec}^{2} \theta=\frac{144}{25}-\frac{169}{25}=-1\)
(v) (b): \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (Using identity)