Given, \(\sqrt { 2 } \sin { \theta } =1\)
\(\sin { \theta } =\frac { 1 }{ \sqrt { 2 } } =\sin { { 45 }^{ ° } } \)
\(\therefore \quad \theta ={ 45 }^{ ° }\)
Now \(\sec ^{ 2 }{ \theta } -cosec^{ 2 }\theta \) = \(\sec ^{ 2 }{ { 45 }^{ ° } } -cosec^{ 2 }{ 45 }^{ ° }\)
\(={ \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 }\)
= 2 - 2
= 0 

LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved.

3 cm

Let AB is pole and BC is its shadow
∴ AB = 6m, BC = 2\(\\ \\ \\ \\ \sqrt { 3 } \) m
In right ΔABC, \(\frac { AB }{ BC } \)=tanፀ
⇒ tanፀ = \(\frac { 6 }{ 2\sqrt { 3 } } \)
⇒ tanፀ = \(\sqrt { 3 } \)
⇒ ፀ = 60^o

Let AB is the tower and C is the point 20 m away from the base of the tower 
∴ BC = 20m, ∠ACB = 45^o

In right ΔABC, tan 45^o = \(\frac{AB}{BC}\)
⇒ I = \(\frac{AB}{20}\) ∴ AB = 20 m

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

Given, \(\tan { \theta } +\sin { \theta } =m\quad \quad \quad ...(i)\)
and \(\tan { \theta } -\sin { \theta } =n\quad \quad ...(ii)\)
On adding Eqs. (i) and (ii), we get
\(2\tan { \theta } =m+n\quad \Rightarrow \tan { \theta } =\frac { m+n }{ 2 } \)
\(\therefore \quad \cot { \theta } =\frac { 1 }{ \tan { \theta } } =\frac { 2 }{ m+n } \quad \quad ...(iii)\)
On subtracting Eq. (ii) from Eq. (i), we get, \(2\sin { \theta } =m-n\)
\(\Rightarrow \sin { \theta } =\frac { m-n }{ 2 } \quad \quad \quad \quad .....(iv)\)
\(\therefore \quad cosec\theta =\frac { 1 }{ \sin { \theta } } =\frac { 2 }{ m-n } \)
We know that, \({ cosec }^{ 2 }\theta -\cot ^{ 2 }{ \theta } =1\)
\({ \left( \frac { 2 }{ m-n } \right) }^{ 2 }-{ \left( \frac { 2 }{ m+n } \right) }^{ 2 }=1\)   [from Eqs. (ii) and (iv)]
\(\Rightarrow \frac { 4 }{ { \left( m-n \right) }^{ 2 } } -\frac { 4 }{ { \left( m+n \right) }^{ 2 } } =1\)
\(\Rightarrow 4\left[ \frac { 1 }{ { \left( m-n \right) }^{ 2 } } -\frac { 1 }{ { \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow 4\left[ \frac { { \left( m+n \right) }^{ 2 }-{ \left( m-n \right) }^{ 2 } }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow 4\left[ \frac { ({ m }^{ 2 }+{ n }^{ 2 }+2mn)-({ m }^{ 2 }+{ n }^{ 2 }-2mn) }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow 4\left[ \frac { 2mn+2mn) }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } \right] =1\)
\(\Rightarrow \frac { 16mn }{ { \left( m-n \right) }^{ 2 }{ \left( m+n \right) }^{ 2 } } =1\)
\(\Rightarrow \frac { 16mn }{ { \left( { m }^{ 2 }-{ n }^{ 2 } \right) }^{ 2 } } =1\)
\(\Rightarrow { \left( { m }^{ 2 }-{ n }^{ 2 } \right) }^{ 2 }=16mn\)
\(\therefore \quad { \left( { m }^{ 2 }-{ n }^{ 2 } \right) }=4\sqrt { mn } \)   [taking positive square roor]
Hence proved. 

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

We have, sec 5A = cosec (A + 30⁰)
\(\Rightarrow\) sec 5A = sec[90⁰- (A + 30⁰)]  \([\because sec({ 90 }^{ 0 }-\theta )=cosec\theta ]\)
\(\Rightarrow\) sec 5A = sec(60⁰ - A)
\(\Rightarrow\) 5A = 60⁰ - A  [\(\therefore\) 5A and (60⁰- A) are acute angles]
\(\Rightarrow\) 6A = 60⁰
\(\therefore\) A = 10⁰ 

45°

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig). We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, by Theorem.
OP \(\perp\) AB
Now AB is a chord of the circle C₁ and OP \(\perp\) AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P see fig. We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then \(\angle\)OQP and \(\angle\)ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore,   \(\Delta\)OQP \(\cong\)\(\Delta\) ORP   (RHS)
This gives  PQ = PR    (CPCT)

Let 'h' m be height of the tower PQ
AB=20 m. Let BQ=x m
In rt. ΔPQB
\(\frac { PQ }{ BQ } \)=tan 60^o  ⇒ \(\frac { h }{ x } =\sqrt { 3 } \Rightarrow \sqrt { 3 } x\) .........(i)
In rt. ΔPQA,
\(\frac { PQ }{ AQ } \)=tan 30^o   ⇒ \(\frac { h }{ 20+x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } \)h=20+x     ....(ii)
From (i) and (ii)
\(\sqrt { 3 } .\sqrt { 3 } \)x=20+x ⇒ x=10
∴ Distance of tower fro A is 20 m+10 m=30 m
Put x=10 in (i), we get
h=\(\sqrt { 3 } \) x 10=1.732 x 10=1.732 m

Let CD = h m be the height of the lighthouse and A, B be two ships on either side of lighthouse, such that their angles of depression are 60° and 45°, respectively.
i.e. ∠XCB = ∠CBD = 45° [alternate angles] and ∠YCA = ∠CAD = 60° [alternate angles]
Given, the distance between the ships A and B is
\(A B=\frac{100(\sqrt{3}+1)}{\sqrt{3}} \mathrm{~m}\)
In right angled ΔBDC,
\(\tan 45^{\circ}=\frac{C D}{B D} \left[\because \tan \theta=\frac{P}{B}\right]\)
\(\Rightarrow 1=\frac{C D}{B D} \left[\because \tan 45^{\circ}=1\right]\)
⇒ BD = CD ......(i)

and in right angled ΔADC,
\(\tan 60^{\circ}=\frac{C D}{A D}\)
\(\Rightarrow \sqrt{3}=\frac{C D}{A D} \left[\because \tan 60^{\circ}=\sqrt{3}\right]\)
\(\Rightarrow C D=\sqrt{3} A D \Rightarrow A D=\frac{C D}{\sqrt{3}}\)
(i) 100 m,
(ii) \(100\left(1+\frac{1}{\sqrt{3}}\right) \mathrm{m}\)

First term
\(=\sin { A } \left( \frac { 1-\cos { A } +1+\cos { A } }{ 1-\sin ^{ 2 }{ A } } \right) =\sin { A } .\frac { 2 }{ \sin ^{ 2 }{ A } } \)
\(=\frac { 2 }{ \sin { A } } =cosecA\quad \quad \left[ \therefore \frac { 1 }{ \sin { A } } =cosecA \right] \)
= Third term
Second term =\(\sqrt { \frac { 1+\cos { A } }{ 1-\cos { A } } } +\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } } \)
\(=\sqrt { \frac { 1+\cos { A } }{ 1-\cos { A } } \times \frac { 1+\cos { A } }{ 1+\cos { A } } } +\sqrt { \frac { 1-\cos { A } }{ 1+\cos { A } } \times \frac { 1-\cos { A } }{ 1-\cos { A } } } \)
\(=\frac { 1+\cos { A } }{ \sqrt { \sin ^{ 2 }{ A } } } +\frac { 1-\cos { A } }{ \sqrt { \sin ^{ 2 }{ A } } } \quad [\because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1]\)
\(=\frac { 1+\cos { A } +1-\cos { A } }{ \sin { A } } \)
= Third term  \([\because \frac { 1 }{ \sin { A } } =cosecA]\)
Hence proved. 

LHS = \(\frac { \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } ++1 }{ (1+\cos { \theta } )\sin { \theta } } \)
\(=\frac { 1+2\cos { \theta } +1 }{ (1+\cos { \theta } )\sin { \theta } } =\frac { 2(\cos { \theta } +1) }{ (1+\cos { \theta } )\sin { \theta } } =2cosec\theta \) 

Given, CD = 6 cm, BD = 8 cm and radius = 4 cm

Join OC, OA and OB.
Let the circle touches the other sides AB and AC at points E and F, respectively.
We know that tangents drawn from an external point to the circle are equal in length.
\(\therefore\) CD = CF = 6 cm            [\(\because\) C is an external point]
BD = BE = 8 cm                [\(\because\) B is an external point]
and AF = AE = x cm (say  [\(\because\) A is an external point]
Area of \(\Delta\)OCB, \(A_1=\frac{1}{2} \times \text { Base } \times \text { Height }\)
\(\begin{aligned} & =\frac{1}{2} \times C B \times O D \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & \quad[\because C B=C D+B D=6+8=14] \end{aligned}\)
Area of \(\Delta\)OCA,
\(\begin{aligned} A_2 & =\frac{1}{2} \times A C \times O F \end{aligned}\)
\(\begin{aligned} =\frac{1}{2}(6+x) \times 4=(12+2 x) \mathrm{cm}^2 \end{aligned}\)
and area of \(\Delta\)OBA,
\(A_3=\frac{1}{2} \times A B \times O E=\frac{1}{2}(8+x) \times 4=(16+2 x) \mathrm{cm}^2\)
Thus, area of  \(\Delta\)ABC
= A₁ + A₂ + A₃ = [28 + (12 + 2x) + (16+ 2x)]
= (56 + 4x) cm²     ...(i)
Now, semi-perimeter of \(\Delta\)ABC=\(\frac{1}{2}\)(AB + BC + CA)
\(\Rightarrow \quad s=\frac{1}{2}(x+8+14+6+x)\)
\(\Rightarrow\) s = (14 + x) cm
Using Heron's formula,
area of \(\Delta\)ABC = \(\begin{aligned} & =\sqrt{s(s-a)(s-b)(s-c)} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(14+x) \times x \times 8 \times 6} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x) 48 x} \end{aligned}\)     ...(ii)
From Eqs. (i) and (ii), we get
\(\sqrt{(14+x) 48 x}=56+4 x=4(14+x)\)
On squaring both sides, we get
(14 + x) 48 x = 4² (14 + x)²
\(\Rightarrow\) 3x = 14 + x
\(\Rightarrow\) 2x = 14
\(\Rightarrow\) x = 7
\(\therefore\)  Length of AC = 6 + x = 6 + 7 = 13 cm
and length of AB = 8 + x = 8 + 7 = 15 cm

In ΔAOB and ΔAOC,
AB=AC, AO=AO and OB=OC
∴ ΔAOB≌∆AOC
∴ ㄥBAO=ㄥCAO
⇒ ㄥCAO=1/2ፀ
In right ΔACO,
\(\frac { AO }{ OC } =cosec\left( \frac { \theta }{ 2 } \right) \)
⇒ \(\frac { AO }{ a } =cosec\left( \frac { \theta }{ 2 } \right) \)
⇒ AO=a.cosec\(\left( \frac { \theta }{ 2 } \right) \).sin¢
∴ Height of centre of balloon is a sin¢.cosec\(\left( \frac { \theta }{ 2 } \right) \).

Let AB be the tower and BC be its shadow.

Given, AB = BC

In right angled △ABC,

\(\tan C=\frac{\text { Perpendicular }}{\text { Base }}=\frac{A B}{B C}=1\)

⇒ tan C = tan45° [∵ tan45° = 1]

∴ C = 45°

A line which intersects circle at two distinct points is known as secant.

Let C₁ and C₂ be the two concentric circles, with centre at O and respective radii being r₁ = 4cm and r₂ = 5cm.
Draw a chord AC of circle C₂, to touch circle C₁ at B.
Join OB.

Here OB 丄 AC [As, tangent at any point of circle is perpendicular to radius through the point of contact]
Thus, in right angled OAB ,we have:
OA² = AB² + OB²     [By using phythagoras theorem]
⇒  52 =  AB² + 4²
⇒  AB² = 25 - 16  = 9
∴ Length of chord AC = 2 AB = 2 x 3 = 6 cm

If Assertion is incorrect but Reason is correct.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(a) (iv) NS and NR are both tangent to the circle
So, \(N S \perp S A\) and \(N R \perp R A\) 
[ \(\therefore\)  Tangent to a circle is perpendicular to the radius through the point of contact ]
\(\therefore \angle N S A=90^{\circ}\) 
(b) (ii) 
\(\angle N S A=90^{\circ} \quad \Rightarrow N A^{2}=N S^{2}+S A^{2}\) [ By Pythagora's Theorem]
\(\Rightarrow N S=\sqrt{N A^{2}-S A^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169-25}\) 
\(=\sqrt{144}=12 \mathrm{~m}\) 
(c) (iiii)
NR = NS = 12m
Tangents drawn from an external point are equal
\(\therefore N R=12 \mathrm{~m}\) 
(d) (iv)
\(\angle S N R+\angle R A S=180^{\circ} \Rightarrow \angle R A S=180^{\circ}-\angle S N R\) 
[\(\because\) The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre]
\(\Rightarrow \angle R A S=180^{\circ}-\theta\) 
(e) (i) 
\(\angle R A S=180^{\circ}-\theta\) [ Calulated bove]
Now, \(\angle N A S=\angle N A R=\frac{1}{2} \angle R A S\) 
[\(\because\) If two tangents are drawn from an external point, then they extend equal angles at the centre]
\(\Rightarrow \angle N A S=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{\theta}{2}\)

(i) (b): \(\angle X A C=45^{\circ}\) 
\(\therefore \quad \angle A C D=45^{\circ}\) [Alternate interior angles]
(ii) (b)
(iii) (c) : \(\text { In } \Delta A C D\)
\(\frac{A D}{D C}=\tan 45^{\circ} \)
\(\Rightarrow \frac{100}{D C}=1 \Rightarrow D C=100 \mathrm{~m}\)
(iv) (d): \(\text { In } \Delta A B D, \frac{A D}{B D}=\tan 30^{\circ}\)
\(\Rightarrow \quad \frac{100}{B D}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow \quad B D=100 \sqrt{3} \mathrm{~m}\)
(v) (a): \(\text { In } \Delta A D C\)
\(\frac{A D}{A C}=\sin 45^{\circ} \Rightarrow \frac{100}{A C}=\frac{1}{\sqrt{2}} \Rightarrow A C=100 \sqrt{2} \mathrm{~m}\)

(i) (b): The person who makes small angle of elevation is more closer to the balloon.
\(\therefore\) Radlra is more closer to the balloon.
(ii) (b): \(\text { In } \Delta E F D, \tan 30^{\circ}=\frac{E D}{D F}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D F} \)
\(\Rightarrow \quad D F=h \sqrt{3} \mathrm{~m}\)
(iii) (a): In \(\Delta\)GCE,
\(\begin{array}{l} \tan 60^{\circ}=\frac{E C}{G C}=\frac{h+4}{D F} \\ \Rightarrow \quad \sqrt{3}=\frac{h+4}{\sqrt{3} h} \Rightarrow 3 h=h+4 \Rightarrow h=2 \end{array}\)
(iv) (c): Height of the balloon from the ground = BE = BC + CD + DE = 2 + 4 + 2 = 8 m
(v) (b)