Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Miscellaneous Exercise 2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Miscellaneous Exercise 2

I. Select the correct option from the given alternatives.

Question 1.

The value of the expression

cos1°. cos2°. cos3° … cos 179° =

(A) -1

(B) 0

(C) \(\frac{1}{\sqrt{2}}\)

(D) 1

Answer:

(B) 0

Explanation:

cos 1° cos 2° cos 3° … cos 179°

= cos 1° cos 2° cos 3° … cos 90°… cos 179°

= 0 …[∵ cos 90° = 0]

Question 2.

\(\frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}+\frac{1+\sec \mathrm{A}}{\tan \mathrm{A}}\) is equal to

(A) 2cosec A

(B) 2 sec A

(C) 2 sin A

(D) 2 cos A

Answer:

(A) 2cosec A

Explanation:

Question 3.

If α is a root of 25cos^{2} θ + 5cos θ – 12 = 0, \(\frac{\pi}{2}\) < α < π, then sin 2α is equal to

(A) \(-\frac{24}{25}\)

(B) \(-\frac{13}{18}\)

(C) \(\frac{13}{18}\)

(D) \(\frac{24}{25}\)

Answer:

(A) \(-\frac{24}{25}\)

Explanation:

25 cos^{2} θ + 5 cos θ – 12 = 0

∴ (5cos θ + 4) (5 cos θ – 3) = 0

∴ cos θ = \(-\frac{4}{5}\) or cos θ = \(\frac{3}{5}\)

Since \(\frac{\pi}{2}\) < α < π,

cos α < 0

∴ cos α = \(-\frac{4}{5}\)

sin^{2} α = 1 – cos^{2} α = 1 – \(\frac{16}{25}=\frac{9}{25}\)

∴ sin α = \(\pm \frac{3}{5}\)

Since \(\frac{\pi}{2}\) < α < π sin α > 0

∴ sin α = 3/5

sin 2 α = 2 sin α cos α

= \(2\left(\frac{3}{5}\right)\left(\frac{-4}{5}\right)=-\frac{24}{25}\)

Question 4.

If θ = 60°, then \(\frac{1+\tan ^{2} \theta}{2 \tan \theta}\) is equal to

(A) \(\frac{\sqrt{3}}{2}\)

(B) \(\frac{2}{\sqrt{3}}\)

(C) \(\frac{1}{\sqrt{3}}\)

(D) \(\sqrt{3}\)

Answer:

(B) \(\frac{2}{\sqrt{3}}\)

Explanation:

Question 5.

If sec θ = m and tan θ = n, then \(\frac{1}{m}\left\{(m+n)+\frac{1}{(m+n)}\right\}\) is equal to

(A) 2

(B) mn

(C) 2m

(D) 2n

Answer:

(A) 2

Explanation:

Question 6.

If cosec θ + cot θ = \(\frac{5}{2}\), then the value of tan θ is

(A) \(\frac{14}{25}\)

(B) \(\frac{20}{21}\)

(C) \(\frac{21}{20}\)

(D) \(\frac{15}{16}\)

Answer:

(B) \(\frac{20}{21}\)

Explanation:

cosec θ + cot θ = \(\frac{5}{2}\) …………….(i)

cosec^{2} θ – cot^{2} θ = 1

∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1

∴ \(\frac{5}{2}\) (cosec θ – cot θ) = 1

∴ cosec θ – cot θ = \(\frac{2}{5}\) …(ii)

Subtracting (ii) from (i), we get

2 cot θ = \(\frac{5}{2}-\frac{2}{5}=\frac{21}{10}\)

∴ cot θ = \(\frac{21}{20}\)

∴ tan θ = \(\frac{20}{21}\)

Question 7.

\(1-\frac{\sin ^{2} \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}-\frac{\sin \theta}{1-\cos \theta}\) equals

(A) 0

(B) 1

(C) sin θ

(D) cos θ

Answer:

(D) cos θ

Explanation:

Question 8.

If cosec θ – cot θ = q, then the value of cot θ is

(A) \(\frac{2 q}{1+q^{2}}\)

(B) \(\frac{2 q}{1-q^{2}}\)

(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

(D) \(\frac{1+q^{2}}{2 q}\)

Answer:

(C) \(\frac{1-\mathrm{q}^{2}}{2 \mathrm{q}}\)

Explanation:

cosec θ – cot θ = q ……(i)

cosec^{2} θ – cot^{2} θ = 1

∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1

∴ (cosec θ + cot θ)q = 1

∴ cosec θ + cot θ = 1/q …….(ii)

Subtracting (i) from (ii), we get

2cot θ = \(\frac{1}{\mathrm{q}}-\mathrm{q}\)

∴ cot θ = \(\frac{1-q^{2}}{2 q}\)

Question 9.

The cotangent of the angles \(\frac{\pi}{3}, \frac{\pi}{4}\) and \(\frac{\pi}{6}\) are in

(A) A.P.

(B) G.P.

(C) H.P.

(D) Not in progression

Answer:

(B) G.P.

Explanation:

Question 10.

The value of tan 1°.tan 2° tan 3° equal to

(A) -1

(B) 1

(C) \(\frac{\pi}{2}\)

(D) 2

Answer:

(B) 1

Explanation:

tan1° tan2° tan3° … tan89°

= (tan 1° tan 89°) (tan 2° tan 88°)

…(tan 44° tan 46°) tan 45°

= (tan 1 ° cot 1 °) (tan 2° cot 2°)

…(tan 44° cot 44°) . tan 45°

…tan(∵ 90° – θ) = cot θ]

= 1 x 1 x 1 x … x 1 x tan 45° =1

II. Answer the following:

Question 1.

Find the trigonometric functions of:

90°, 120°, 225°, 240°, 270°, 315°, -120°, -150°, -180°, -210°, -300°, -330°

Solution:

Angle of measure 90° :

Let m∠XOA = 90°

Its terminal arm (ray OA)

intersects the standard, unit circle at P(0, 1).

∴ x = 0 and y = 1

sin 90° = y = 1

cos 90° = x = 0

tan 90° = \(\frac{y}{x}=\frac{1}{0}\), which is not defined

cosec 90° = \(\frac{1}{y}=\frac{1}{1}\) = 1

sec 90° = \(\frac{1}{x}=\frac{1}{0}\), which is not defined

cot 90° = \(\frac{x}{y}=\frac{0}{1}\) = 0

Angle of measure 120° :

Let m∠XOA =120°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Since point P lies in the 2nd quadrant, x < 0, y > 0

[Note: Answer given in the textbook of tan 120° is \(\frac{-1}{\sqrt{3}}\) and cot 120° is \(-\sqrt{3}\). However, as per our \(-\sqrt{3}\) calculation the answer of tan 120° is \(-\sqrt{3}\) and cot 120° is \(-\frac{1}{\sqrt{3}}\)

Angle of measure 225° :

Let m∠XOA = 225°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 45° – 45° – 90° triangle.

OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 240° :

Let m∠XOA = 240°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure 270° :

Let m∠XOA = 270°

Its terminal arm (ray OA) intersects the standard unit circle at P(0, – 1).

x = 0 andy = – 1

sin 270° = y = -1

cos 270° = x = 0

tan 270° = \(\frac{y}{x}\)

Angle of measure 315° :

Let m∠XOA = 315°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 45° – 45° – 90° triangle.

OP = 1

[Note: Answer given in the textbook of cot 315° is 1. However, as per our calculation it is -1.]

Angle of measure (-120°):

Let m∠XOA = – 120°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (-150°) :

Let m∠XOA = – 150°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Angle of measure (-180°):

Let m∠XOA = – 180°

Its terminal arm (ray OA) intersects the standard unit circle at P(- 1, 0).

∴ x = – 1 andy = 0

sin (-180°) = y = 0

cos (-180°) = x

= -1

Angle of measure (- 210°):

Let m∠XOA = -210°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Angle of measure (- 300°):

Let m∠XOA = – 300° Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

ΔOMP is a 30° – 60° – 90° triangle.

OP = 1

Since point P lies in the 1st quadrant, x>0,y>0

x = OM = \(\frac{1}{2}\) and

y = PM = \(\frac{\sqrt{3}}{2}\)

Angle of measure (- 330°):

Let m∠XOA = – 330°

Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).

Draw seg PM perpendicular to the X-axis.

∴ ΔOMP is a 30° – 60° – 90° triangle.

OP= 1

Since point P lies in the 1st quadrant, x > 0, y > 0

∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Question 2.

State the signs of:

i. cosec 520°

ii. cot 1899°

iii. sin 986°

Solution:

i. 520° =360° + 160°

∴ 520° and 160° are co-terminal angles.

Since 90° < 160° < 180°,

160° lies in the 2nd quadrant.

∴ 520° lies in the 2nd quadrant,

∴ cosec 520° is positive.

ii. 1899° = 5 x 360° + 99°

∴ 1899° and 99° are co-terminal angles.

Since 90° < 99° < 180°,

99° lies in the 2nd quadrant.

∴ 1899° lies in the 2nd quadrant.

∴ cot 1899° is negative.

iii. 986° = 2x 360° + 266°

∴ 986° and 266° are co-terminal angles.

Since 180° < 266° < 270°,

266° lies in the 3rd quadrant.

∴ 986° lies in the 3rd quadrant.

∴ sin 986° is negative.

Question 3.

State the quadrant in which 6 lies if

i. tan θ < 0 and sec θ > 0

ii. sin θ < 0 and cos θ < 0

iii. sin θ > 0 and tan θ < 0

Solution:

i. tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0

sec θ is positive in 1st and 4th quadrants.

∴ θ lies in the 4th quadrant.

ii. sin θ < 0

sin θ is negative in 3rd and 4th quadrants, cos θ < 0

cos θ is negative in 2nd and 3rd quadrants.

.’. θ lies in the 3rd quadrant.

iii. sin θ > 0

sin θ is positive in 1st and 2nd quadrants, tan θ < 0

tan θ is negative in 2nd and 4th quadrants.

∴ θ lies in the 2nd quadrant.

Question 4.

Which is greater?

sin (1856°) or sin (2006°)

Solution:

1856° = 5 x 360° + 56°

∴ 1856° and 56° are co-terminal angles.

Since 0° < 56° < 90°, 56° lies in the 1st quadrant.

∴ 1856° lies in the 1st quadrant,

∴ sin 1856° >0 …(i)

2006° = 5 x 360° + 206°

∴ 2006° and 206° are co-terminal angles.

Since 180° < 206° < 270°,

206° lies in the 3rd quadrant.

∴ 2006° lies in the 3rd quadrant,

∴ sin 2006° <0 …(ii)

From (i) and (ii),

sin 1856° is greater.

Question 5.

Which of the following is positive?

sin(-310°) or sin(310°)

Solution:

Since 270° <310° <360°,

310° lies in the 4th quadrant.

∴ sin (310°) < 0

-310° = -360°+ 50°

∴ 50° and – 310° are co-terminal angles.

Since 0° < 50° < 90°, 50° lies in the 1st quadrant.

∴ – 310° lies in the 1st quadrant.

∴ sin (- 310°) > 0

∴ sin (- 310°) is positive.

Question 6.

Show that 1 – 2sin θ cos θ ≥ 0 for all θ ∈ R.

Solution:

1 – 2 sin θ cos θ

= sin^{2} θ + cos^{2} θ – 2sin θ cos θ

= (sin θ – cos θ)^{2} ≥ 0 for all θ ∈ R

Question 7.

Show that tan^{2} θ + cot^{2} θ ≥ 2 for all θ ∈ R.

Solution:

Question 8.

If sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\), then find the values of cos θ, tan θ in terms of x and y.

Solution:

Given, sin θ = \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\)

we know that

cos^{2}θ = 1 – sin^{2 }θ

[Note: Answer given in the textbook of cos θ = \(\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = \(. However, as per our calculation the answer of cos θ = ± [latex]\frac{2 x y}{x^{2}+y^{2}}\) and tan θ = ± \(\frac{x^{2}-y^{2}}{2 x y}\). ]

Question 9.

If sec θ = \(\sqrt{2}\) and \(\frac{3 \pi}{2}\) < θ < 2π, then evaluate \(\frac{1+\tan \theta+{cosec} \theta}{1+\cot \theta-{cosec} \theta}\)

Solution:

Given sec θ = \(\sqrt{2}\)

We know that,

tan^{2} θ = sec^{2} θ – 1

= (\(\sqrt{2}\)) – 1

= 2 – 1 = 1

∴ tan θ = ±1

Since \(\frac{3 \pi}{2}\) < θ < 2π

θ lies in the 4th quadrant.

∴ tan θ < 0

∴ tan θ = -1

Question 10.

Prove the following:

i. sin^{2}A cos^{2} B + cos^{2}A sin^{2}B + cos^{2}A cos^{2}B + sin^{2}A sin^{2}B = 1

Solution:

L.H.S. = sin^{2}A cos^{2} B + cos^{2}A sin^{2}B + cos^{2}A cos^{2}B + sin^{2}A sin^{2}B

= sin^{2}A (cos^{2} B + sin^{2} B) + cos^{2} A (sin^{2} B + cos^{2} B)

= sin^{2}A(1) + cos^{2}A(1)

= 1 = R.H.S.

ii. \(\frac{(1+\cot \theta+\tan \theta)(\sin \theta-\cos \theta)}{\sec ^{3} \theta-{cosec}^{3} \theta}=\sin ^{2} \theta \cos ^{2} \theta\)

Solution:

iii. L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}=2\left(\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}\right)\)

Solution:

L.H.S. = \(\left(\tan \theta+\frac{1}{\cos \theta}\right)^{2}+\left(\tan \theta-\frac{1}{\cos \theta}\right)^{2}\)

= (tanθ + secθ)^{2} + (tanθ – secθ)^{2}

= tan^{2} θ + 2 tan θ sec θ + sec^{2} θ

+ tan^{2} θ – 2 tan θ sec θ +.sec^{2} θ

= 2(tan^{2} θ + sec^{2} θ)

iv. 2.sec^{2} θ – sec^{4} θ – 2.cosec^{2} θ + cosec^{4} θ = cot^{4} θ – tan^{4} θ

Solution:

LHS.

= 2.sec^{2} θ – sec^{4} θ – 2.cosec^{2} θ + cosec^{4} θ = = 2 sec^{2} θ – (sec^{2} θ)^{2} – 2cosec^{2} θ + (cosec^{2} θ)^{2}

= 2(1+ tan^{2} θ) – (1+ tan^{2} θ)^{2} – 2(1+ cot^{2} θ)

+ (1+ cot^{2} θ)^{2}

= 2 + 2tan^{2} θ – (1 + 2tan^{2} θ + tan^{4} θ)

– 2 – 2cot^{2} θ + 1 + 2cot^{2} θ + cot^{4} θ

= 2 + 2.tan^{2} θ – 1 – 2 tan^{2} θ – tan^{4} θ – 2

– 2 cot^{2} θ + 1 + 2 cot^{2} θ + cot^{4} θ

= cot^{4} θ – tan^{4} θ = R.H.S.

v. sin^{4} θ + cos^{4} θ = sin^{4} θ + cos^{4} θ

Solution:

L.H.S. = sin^{4} θ + cos^{4} θ

= (sin^{2} θ)^{2} + (cos^{2} θ)^{2} = (sin^{2} θ + cos^{2} θ)^{2} – 2sin^{2} θ cos^{2} θ

… [ v a^{2} + b^{2} = (a + b)^{2} – 2ab]

= 1 – 2sin^{2} θ cos^{2} θ

= R.H.S.

vi. 2(sin^{6} θ + cos^{6} θ) – 3(sin^{4} θ + cos^{4} θ) + 1 = 0

L.H.S =

2(sin^{6} θ + cos^{6} θ) – 3(sin^{4} θ + cos^{4} θ) + 1=0

= sin^{6} θ + cos^{6} θ

= (sin^{2} θ)^{3} + (cos^{2} θ)^{3} = (sin^{2} θ + cos^{2} θ)^{3}

– 3 sin^{2} θ cos^{2} θ (sin2 0 + cos2 0)

…[••• a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b)]

= (1)^{3} – 3 sin^{2} θ cos^{2} θ(1)

= 1-3 sin^{2} θ cos^{2} θ sin^{4} θ + cos^{4} θ

= (sin^{2} θ)^{2} + (cos^{2} θ)^{2} = (sin^{2} θ + cos^{2} θ)^{2} – 2 sin^{2} θ cos^{2} θ

…[Y a^{2} + b^{2} = (a + b)^{2} – 2ab]

= 1-2 sin^{2} θ cos^{2} θ

L.H.S.= 2(sin^{6} θ + cos^{6} θ) – 3(sin^{4} θ + cos^{4} θ) + 1

= 2(1-3 sin^{2} θ cos^{2} θ) -3(1 – 2 sin^{2} θ cos^{2} θ) + 1

= 2-6 sin^{2} θ cos^{2} θ – 3 + 6 sin^{2} θ cos^{2} θ + 1 = c

= R.H.S.

vii. cos^{4} θ – sin^{4} θ + 1 = 2cos^{2}θ

L.H.S. = cos^{4} θ – sin^{4} θ + 1

= (cos^{2} θ)^{2} – (sin^{2} θ)^{2} + 1 = (cos^{2}θ + sin^{2}θ) c(os^{2} θ – sin^{2}θ) +1

= (1) (cos^{2}θ – sin^{2}θ) + 1 = cos^{2} θ + (1 – sin^{2}θ)

= cos^{2} θ + cos^{2}θ = 2cos^{2}θ = R.H.S.

viii. sin^{4}θ + 2sin^{2}θ cos^{2}θ = 1 – cos^{4}θ

L.H.S. = sin^{4}θ + 2sin^{2}θ cos^{2}θ = sin^{2}θ(sin^{2}θ + 2cos^{2}θ)

= (sin^{2}θ) (sin^{2}θ + cos^{2}θ + cos^{2}θ) = (1 – cos^{2}θ) (1 + cos^{2}θ)

= 1 – cos^{4}θ = R.H.S.

ix. \(\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}+\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta-\cos \theta}=2\)

Solution:

= (sin^{2} θ + cos^{2} θ – sin θ cos θ) + (sin^{2} θ + cos^{2} θ + sinθ cosθ)

= 2 (sin^{2} θ + cos^{2} θ)

= 2(1)

= 2 = R.H.S.

x. tan^{2} θ – sin^{2} θ = sin^{4} θ sec^{2} θ

Solution:

L.H.S. = tan^{2} θ – sin^{2} θ

= \(\frac{\sin ^{2} \theta}{\cos ^{2} \theta}\) – sin^{2}θ

= sin^{2} θ (\(\frac{1}{\cos ^{2} \theta}-1 \))

= \(\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}\)

= (sin^{2} θ) (sin^{2} θ)sec^{2} θ

= sin^{4} θ sec^{2} θ

= R.H.S

xi. (sinθ + cosecθ)^{2} + (cos θ + see θ)^{2 }= tan^{2} θ + cot^{2} θ + 7

Solution:

L.H.S. = (sinθ + cosecθ)^{2} + (cos θ + see θ)^{2}

= sin ^{2} θ + cosec^{2} θ + 2sinθ cosec θ

+ cos^{2} θ + sec^{2} θ + 2sec0 cos0

= (sin^{2} θ + cos^{2} θ) + cosec^{2} θ + 2 + sec^{2} θ + 2

= 1 + (1 + cot^{2} θ) + 2 + (1 + tan^{2} θ) + 2 = tan^{2} θ + cot^{2} θ + 7

= R.H.S.

xii. sin^{8}θ – cos^{8}θ = (sin^{2} θ – cos^{2} θ) (1 – 2sin^{2} θ cos^{2} θ)

Solution:

L.H.S. = sin^{8}θ – cos^{8}θ

= (sin^{4}θ)^{2} – (cos^{4}θ)^{2}

= (sin^{4}θ – cos^{4}θ) (sin^{4}θ + cos^{4}θ)

= [(sin^{2} θ)^{2} – (cos^{2} θ)^{2} ]

. [(sin^{2} θ)^{2} + (cos^{2} θ)^{2} ]

= (sin^{2} θ + cos^{2} θ) (sin^{2} θ – cos^{2} θ). [(sin^{2} θ + cos^{2} θ)^{2} – 2sin^{2} θ.cos^{2} θ] …[Y a^{2} + b^{2} = (a + b)^{2} – 2ab]

= (1) (sin^{2} θ – cos^{2} θ) (1^{2} – 2sin^{2} θ cos^{2} θ)

= (sin^{2} θ – cos^{2} θ) (1 – 2sin^{2} θ cos^{2} θ)

= R.H.S.

xiii. sin^{6}A + cos^{6}A = 1 – 3 sin^{2}A + 3sin^{4}A

Soluiton:

L.H.S. = sin^{6}A + cos^{6}A

= (sin^{2} A)^{3} + (cos^{2} A)^{3}

= (sin^{2} A + cos^{2} A)^{3}

– 3sin^{2}A cos^{2}A(sin^{2} A + cos^{2} A)

…[ a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b)]

= 1^{3} – 3sin^{2}A cos^{2}A (1)

= 1 – 3sin^{2}A cos^{2}A

= 1 – 3 sin^{2}A (1 – sin^{2}A)

= 1 – 3 sin^{2}A + 3sin^{4}A

= R.H.S.

xiv. (1 + tanA tanB)^{2} + (tanA – tanB)^{2} = sec ^{2}A sec^{2}B

Solution:

L.H.S. = (1 + tanA tanB)^{2} + (tanA – tanB)^{2}

= 1 + 2tanA tanB + tan^{2}A tan^{2} + tan^{2} A- 2tanA tanB + tan^{2}B

= 1 + tan^{2}A + tan^{2} B + tan^{2}A tan^{2}B

= 1(1+ tan^{2}A) + tan^{2} B(1 + tan^{2}A)

= (1 + tan^{2}A) (1 + tan^{2}B)

= sec^{2}A sec^{2}B = R.H.S.

xv. \(\frac{1+\cot \theta+{cosec} \theta}{1-\cot \theta+{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-1}{\cot \theta-{cosec} \theta+1}\)

Solution:

We know that cosec^{2}θ – cot^{2} θ = 1

∴ (cosec θ – cot θ) (cosec θ + cot θ) = 1

xvi. \(\frac{\tan \theta+\sec \theta-1}{\tan \theta+\sec \theta+1}=\frac{\tan \theta}{\sec \theta+1}\)

Solution:

We know that

tan^{2}θ = sec^{2} θ – 1

∴ tan θ. tanθ = (sec θ + 1)(sec θ – 1)

xvii. \(\frac{{cosec} \theta+\cot \theta-1}{{cosec} \theta+\cot \theta+1}=\frac{1-\sin \theta}{\cos \theta}\)

Solution:

We know that,

cot^{2} θ = cosec^{2} θ – 1

∴ cot θ . cot θ = (cosec θ + 1)(cosec θ – 1)

Alternate Method:

xviii. \(\frac{{cosec} \theta+\cot \theta+1}{\cot \theta+{cosec} \theta-1}=\frac{\cot \theta}{{cosec} \theta-1}\)

solution:

We know that,

cot^{2} θ = cosec^{2} θ – 1

∴ cot θ.cot θ = (cosec θ + 1) (cosec θ – 1)