∵ tan A = cot B
tan A = tan (90° – B)
A = 90° – B
A + B = 90°. Proved

Let the first term of the AP be a , and the common difference be d . It is given that the \(2^{\text {nd }}\) term is 13 ie  and also that the \(5^{\text {th }}\) term is 25 , i.e.
\(t_5=a+4 d\)
Solving for d, we get \((a+4 d)-(a+d)=25-13=12\)
Thus, \(3 \mathrm{~d}=12\)
Or, d = 4.
Hence, a=13-4=9. To find the 7th term, we need to find the value of a+6 d i.e. 9+6(4)=33.
Thus the 7th term of the given arithmetic progression is 33 

6

\(sin\ \alpha =\frac { 1 }{ 2 } and\ cos\ \beta =\frac { 1 }{ 2 }\)
\( \\ \Rightarrow d sin\ \alpha =sin\ 30°\left[ \because \ sin\ 30°=\frac { 1 }{ 2 } \right] \ \)
\(\ and\ cos\ \beta \ =\ cos\ 60°\ \left[ \because \ cos\ 60°=\frac { 1 }{ 2 } \right] \)
\(\\ \Rightarrow \ \alpha \ =\ 30°\ and\ \ beta \ = \ 60°\)
\(\\ \therefore \ \alpha \ +\ \beta \ =\ 30°+60°=90°\)

Let AB is tower and C is a point on the ground such that BC = 100 m
In right \(\Delta\)ABC,
\(\frac { AB }{ BC } =\tan { { 45 }^{ o } } \)
\(\Rightarrow\)  \(\frac { AB }{ 100 } =1\)
\(\Rightarrow\)   AB = 100 m

Let AB is tower and BC is its shadow
∴ AB = 20 m and BC = 20\(\sqrt { 3 } \)
In right ΔABC, \(\frac { AB }{ BC } \) = tanፀ
⇒ tanፀ = \(\frac { 20 }{ 20\sqrt { 3 } } \)
⇒ tanፀ = \(\frac { 1 }{ \sqrt { 3 } } \)
⇒ ፀ = 30^o

Here, a = 11 and d = 8 - 11 = -3
Assume that, -150 be the nth term of the given AP.
We know that the nth term of an AP is
a_n = a + (n - 1)d
\(\Rightarrow\) -150 = 11 + (n - 1) (-3)
\(\Rightarrow n-1=\frac{161}{3}\)
\(\Rightarrow n=\frac{161}{3}+1=\frac{164}{3}\)
But n should be a positive integer.
So, - 150 is not a term of the given AP.

Given, \(\sec { \theta } =\frac { 13 }{ 12 } \)
\(\Rightarrow \quad \frac{P R}{P Q}=\frac{13}{12} \quad\left[\because \sec \theta=\frac{H}{B}\right]\)
Let, PR = 13k and PQ = 12k
Where, k is any positive integer.

In right angled \(\triangle PQR,\)
PQ² + RQ² = PR²    
[By using Pythagoras theorem]
\(\Rightarrow\) RQ² + (12k)² =( 13k)²
\(\Rightarrow\) RQ² + 144k² = 169k²
\(\Rightarrow\) RQ² = 169k² - 144k² = 25k²
\(\Rightarrow\) PQ - 5k [using positive square for since, side cannot be negative]
\(\sin { \theta } =\frac { P }{ H } =\frac { 5k }{ 13k } =\frac { 5 }{ 13 } \)
\(\cos { \theta } =\frac { B }{ H } =\frac { 12k }{ 13k } =\frac { 12 }{ 13 } \)
\(\begin{aligned} & \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{5 / 13}{12 / 13}=\frac{5}{13} \times \frac{13}{12}=\frac{5}{12} \\ \end{aligned}\)
\(\begin{aligned} & \operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{13}{5} \text { and } \cot \theta=\frac{1}{\tan \theta}=\frac{12}{5} \end{aligned}\)

LHS = \(\frac { \tan { \theta } }{ 1-\cot { \theta } } +\frac { \cot { \theta } }{ 1-\tan { \theta } } =\frac { \frac { \sin { \theta } }{ \cos { \theta } } }{ 1-\frac { \cos { \theta } }{ \sin { \theta } } } +\frac { \frac { \cos { \theta } }{ \sin { \theta } } }{ 1-\frac { \sin { \theta } }{ \cos { \theta } } } \) \(\left[ \because \tan { A } =\frac { \sin { A } }{ \cos { A } } ,\cot { A } =\frac { \cos { A } }{ \sin { A} } \right] \)
\(=\frac { \frac { \sin { \theta } }{ \cos { \theta } } }{ \frac { \sin { \theta } -\cos { \theta } }{ \sin { \theta } } } +\frac { \frac { \cos { \theta } }{ \sin { \theta } } }{ \frac { \cos { \theta } -\sin { \theta } }{ \cos { \theta } } } \)
\(=\frac { \sin { \theta } }{ \cos { \theta } } \times \frac { \sin { \theta } }{ \sin { \theta } -\cos { \theta } } +\frac { \cos { \theta } }{ \sin { \theta } } \times \frac { \cos { \theta } }{ \cos { \theta } -\sin { \theta } } \)
\(=\frac { \sin ^{ 2 }{ \theta } }{ \cos { \theta } (\sin { \theta } -\cos { \theta } ) } -\frac { \cos ^{ 2 }{ \theta } }{ \sin { \theta } (\sin { \theta } -\cos { \theta } ) } \)
\(=\frac { \sin ^{ 3 }{ \theta } -\cos ^{ 3 }{ \theta } }{ \cos { \theta } \sin { \theta } (\sin { \theta } -\cos { \theta } ) } \)
\(=\frac { (\sin { \theta } -\cos { \theta } )(\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +\sin { \theta } \cos { \theta } ) }{ \cos { \theta } \sin { \theta } (\sin { \theta } -\cos { \theta } ) } \)   \(\left[ \because { x }^{ 3 }-{ y }^{ 3 }=(x-y)({ x }^{ 2 }+{ y }^{ 2 }+xy) \right] \)
\(=\frac { 1+\sin { \theta } \cos { \theta } }{ \cos { \theta } \sin { \theta } } \quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)
\(=\frac { 1 }{ \cos { \theta } \sin { \theta } } +\frac { \sin { \theta } \cos { \theta } }{ \sin { \theta } \cos { \theta } } =\frac { 1 }{ \cos { \theta } \sin { \theta } } +1\)
\(=1+\sec { \theta } cosec\theta \quad \left[ \because \frac { 1 }{ \sin { \theta } } =cosec\theta \quad and\quad \frac { 1 }{ \cos { \theta } } =\sec { \theta } \right] \)
= RHS
Hence proved.

Let AB=60 m is height of building and CD is lamp post
(i) In rt ΔABD,  \(\frac { AB }{ BD } \)=tan60⁰
⇒ \(\frac { 60 }{ BD } =\sqrt { 3 } \Rightarrow \frac { 60 }{ \sqrt { 3 } } \)=BD
⇒ BD=\(\frac { 60\times \sqrt { 3 } }{ 3 } =20\sqrt { 3 } \)m
⇒ BD=20 x 1.732=34.64 m
(ii) In rt. ΔAEC, \(\frac { AE }{ EC } \)=tan30⁰
⇒ \(\frac { AE }{ 20\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } \)            [∵ EC=BD]
⇒ AE=20 m
and EB=AB-AE=60-20=40 m
Also EB=CD
⇒ CD=40 m
∴ Height of lamp post =40 m.

T_p = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}=a+(p-1)d\)                    ...(i)
T_q = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}a+(q-1)d\)                         ...(ii)
Subtracting (ii) from (i), we get
\({p-q\over pq}=(p-q)d\Rightarrow d={1\over pq}\)
Putting d = \({1 \over pq}\) in (i), we have
\({1\over q}=a+{(p-q)\over pq}\Rightarrow a={1\over pq}\)
\(\therefore\)     S_pq = \({pq\over2}[2a+(pq-1)d]\)
or,      S_pq = \({pq\over2}\left[ {{2\over pq}+{(pq-1)\over pq}} \right]\)
\(={1\over 2}(pq+1)\)

\(\because\)         S₆ = 36 and S₁₆ = 256.
\(\Rightarrow\)   S₆ = \({6\over2}\) [ 2a + 5d ]
   [ \(\because\) S_n = \({n\over2} [ 2a + ( n - 1)d ]\)
\(\Rightarrow\)    S₆ = 3 ( 2a + 5d )
\(\Rightarrow\)    \({36\over3}\) = 2a + 5d 
\(\Rightarrow\)    12 = 2a + 5d                       ...(i)
and   S₁₆ = \({16\over2}[2a+15d]\)
\(\Rightarrow\)       \({256\over8}=2a+15d\)
\(\Rightarrow\)   32 = 2a + 15d                     ...(ii)
Subtracting (i) and (ii), 2n + 5d = 12
2a + 15d = 32
-    -           -    
     -10d = -20   \(\Rightarrow\)  d = 2
\(\therefore\)  From (i), 12 = 2a + 5(2)
12 - 10 = 2a  \(\Rightarrow\)  2a = 2 \(\Rightarrow\) a = 1
Hence   S₁₀ = \({10\over2}[2a+9d]\)
= 5 ( 2 x 1 +  9 x 2  )
= 5 ( 2 + 18 )
\(\Rightarrow\)   S₁₀ = 5 x 20 = 100.

Let BC = x m be the width of the canal and AB = h m be the height of the tower.
Given, \(\angle\)ACB = 60\(\unicode{xb0} \) and  \(\angle\)ADB = 30\(\unicode{xb0} \)
In right angled \(\Delta\)ABD, 
\(\tan 30^{\circ}=\frac{P}{B}=\frac{A B}{D B}\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D C+C B}\)

\(\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { and } D B=D C+C B\right]\)
\(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{20+x} \Rightarrow 20+x=\sqrt{3} h\)  ...(i)
and in right angled \(\Delta\)ABC,
\(\tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]\)
\(\Rightarrow \quad h=\sqrt{3} x\)                                      ..(ii)
On putting h = \(\sqrt{3}\)x in Eq. (i), we get
20 + x = \(\sqrt{3}\) (\(\sqrt{3}\)x)
\(\Rightarrow\) 20 + x = 3x
\(\Rightarrow\) 2x = 20 \(\Rightarrow\) x = 10 m
On putting x = 10 m in Eq. (ii), we get h = \(\sqrt{3}\)(10)
\(\Rightarrow\) h = 10\(\sqrt{3}\) m
Hence, the height of the tower is 10\(\sqrt{3}\) m and width of the canal is 10 m.

Trigonometric ratio involving AB, BC, OD, OA and angles is tanθ. [Refer AB, BC, OA and OD from the figure.]
Distance travelled by the balloon OB = AB - OA
From the figure, OD = BC, and it can be calculated as
88.2 m - 1.2 m = 87 m --- (1)
In ΔAOD,
tan 60° = OD/OA
√3 = 87/OA
OA = 87 / √3
= 87 × √3 / √3 × √3
= (87 × √3) / 3
= 29√3 m

\(\cos { \theta } +\sin { \theta } =\sqrt { 2 } \cos { \theta } \)
\(\Rightarrow \quad \sin { \theta } =\cos { \theta } \left( \sqrt { 2 } -1 \right) \)
\(\Rightarrow \quad \sin { \theta } =\frac { \cos { \theta } \left( \sqrt { 2 } -1 \right) \left( \sqrt { 2 } +1 \right) }{ \left( \sqrt { 2 } +1 \right) } \)
\(\Rightarrow \quad \sin { \theta } =\frac { \cos { \theta } \left( 2-1 \right) }{ \sqrt { 2 } +1 } \)
\(\Rightarrow \quad \left( \sqrt { 2 } +1 \right) \sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \sqrt { 2 } \sin { \theta } +\sin { \theta } =\cos { \theta } \)
\(\Rightarrow \quad \cos { \theta } -\sin { \theta } =\sqrt { 2 } \sin { \theta } \)

S_n = 3n² - 4n
S₁ = 3(1)² - 4(1) = - 1
S₂ = 3(2)²- 4(2) = 4
a₁ = S₁ =- 1
a₂ = S₂- S₁ = 4 - (- 1) = 5
d = a₂ - a₁ = 5 - (-1) = 6
A.P. is - 1, 5, 11,............
a₁₂ = a + 11d
=-1 +11 \(\times\) 6
= 65.

since, sin (A - B) = \(\frac{1}{2}\), therefore, A - B = 30°              (1)
Also, since cos (A + B) = \(\frac{1}{2}\), therefore, A + B = 60°   (2)
Solving (1) and (2), we get : A = 45° and B = 15°.

32

Let AB is Ist pole and CD is IInd pole.

CD = 24 m and BD = 15 m
 AE is horizontal line.
In rectangle
 ABDE,
 AE = BD = 15 m
Let CE = x m.
In right \(\Delta \) CEA,
\(\frac { CE }{ AE } =tan{ 30 }^{ o }\Rightarrow \frac { x }{ 15 } =\frac { 1 }{ \sqrt { 3 } } \)m

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Let a man is standing on the deck of a ship at point A such that AB = 10 m and let CD be the hill.
Then, ∠EAD = 60° and ∠CAE = ∠BCA = 30° [alternate angles]
Let BC = x m = AE and DE = h m

In right angled ΔAED, 
\(\tan 60^{\circ}=\frac{P}{B}=\frac{D E}{E A}=\frac{h}{x}\)
In right angled ΔABC,
\(\tan 30^{\circ}=\frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{x}\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]\)
distance of the hill from the ship is 10\(\sqrt3\) m and height of the hill is 40 m.

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

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

(i) (b) Let AB be the monument of height 42 m and C is the point where they are standing such that BC = 42 m.


Now, in \(\Delta\)ABC,
\(\begin{aligned} & \tan \theta=\frac{A B}{B C} \Rightarrow \tan \theta=\frac{42}{42}=1 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \tan \theta & =1 \quad \Rightarrow \quad \theta=45^{\circ} \end{aligned}\)
(ii) (a) In \(\Delta\)ABC,

\(\begin{aligned} \tan 60^{\circ} & =\frac{A B}{B C} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \sqrt{3} & =\frac{42}{B C} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad B C & =\frac{42}{\sqrt{3}}=\frac{42 \sqrt{3}}{3}=14 \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} =14 \times 1.732=24.248=24.25 \mathrm{~m} \end{aligned}\)
(iii) (a) Let AB be the height of the tower.

Then, in \(\Delta\)ABC, tan 60°\(=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{h}{20}\)
\(\Rightarrow\)  h = 20\(\sqrt3\) m
(b) Let h and x be the height and length of shadow of the vertical tower.

Then, in \(\Delta\)ABC,
\(\begin{array}{rlrl} \tan \theta & =\frac{A B}{B C} \Rightarrow \tan \theta=\frac{h}{x} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & & \tan \theta & =1 \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \theta & =45^{\circ} & {[\because h: x=1: 1]} \end{array}\)
(v) (c) The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal, when it is below the horizontal level.

(i) (a): Let the production during first year be a and let d be the increase in production every year. Then,
a₆ = 16000 \(\Rightarrow\)a + 5d = 16,000     (i)
and a₉ = 22600 \(\Rightarrow\)a + 8d = 22600  (ii)
On substracting (i) from (ii) , we get
3d = 6600 \(\Rightarrow\) d = 2200
Putting d = 2200 in (i) we get,
a + 5 x 2200 = 16000
\(\Rightarrow\) a + 11000 = 16000 \(\Rightarrow\) a = 16000 - 11000 = 5000
Thus , a = 5000 and d = 2200
Production during first year, a = 5000.
(ii) (c): Production during 8th year is given by a₈
= (a + 7d)
= (5000 + 7(2200))
= (5000 + 15400)
= 20400.
(iii) (a): a₂ = (a + d)
= (5000 + 2200) = 7200.
a₃ = (a2 + d)
= 7200 + 2200 = 9400.
Production during first 3 years = 5000 + 7200 + 9400 = 21600
(iv) (c): a_n = 5000 + (n – 1)2200 = 29200
(n – 1)2200 = 29200 – 5000 = 24200
⇒ n – 1  = 11
⇒ n = 12
(v) (d): a₄ = (a + 3d)
= (5000 + 3(2200)) = 5000 + 6600 = 11600.
a₇ = (a6 + d)
= 16000 + 2200 = 18200.
Difference = 18200 – 11600 = 6600

\(\because \Delta\)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)