Given: CD = 90 meter;
∠KBD = 30° & ∠CAD = 60°
BD × cos30° = BK
AD × cos60° = AC
BD × sin30° = DK
AD × sin60° = DC
BK = AC
BD × cos30° = AD × cos60°
√ 3 × BD = AD ____________ (1)
Now, DC = DK + KC
DC - DK = AB [As, KC = AB]
(AD × sin60°) - (BD × sin30°) = AB
Using equation (1)
($$\sqrt{3}$$ ×BD×$$\frac{\sqrt{3}}{2}$$)−BD2=AB$\left(\sqrt{3}\times BD\times \frac{\sqrt{3}}{2}\right)-\frac{BD}{2}=A\mathrm{B<br>}$ AB = BD _____________ (2)
Now, AD × sin 60° = DC = 90 meter
AD=$$\frac{180}{\sqrt{3}}$$$\mathrm{<br>}$From (1)
BD=AD/$$\sqrt{3}$$
$<span\; class="mi">=180/\$\$\backslash sqrt\{3\}\; \backslash times\; \backslash sqrt\{3\}\; \$\$\; =\; </span>60mete\mathrm{r<br>}$From (2) AB = 60 meter
Option B is the correct answer.

$$u_6 = cos^6 α + sin^6 α$$
$$2 u_6= 2cos^6 α + 2sin^6 α$$
$$u_4 = cos^4 α + sin^4 α$$
$$3 u_4= 3cos^4 α + 3sin^4 α$$
$$2 u_6 - 3 u_4 = (2cos^6 α + 2sin^6 α) - (3cos^4 α + 3sin^4 α)$$
$$2 cos^6 α = 2(1- sin^2 α)^3 = 2(1 - sin^3α- 3sin^2α+ 3 sin^4 α )$$
$$3 cos^4 α = 3(1- sin^2 α)^2 = 3(1 + sin^4α- 2sin^2α)$$
$$2 cos^6 α - 3 cos^4 α = 2(1 - sin^3α- 3sin^2α+ 3 sin^4 α ) - 3(1 + sin^4α- 2sin^2α) =3sin^4α - 2sin^6α-1$$
$$2 cos^6 α - 3 cos^4 α=2sin^4α - 3sin^6α-1$$
$$2 u_6 - 3 u_4 +1= (2cos^6 α + 2sin^6 α) - (3cos^4 α + 3sin^4 α) +1= 3sin^4α - 2sin^6α - 3sin^4α + 2sin^6α-1+1 $$
$$2 u_6 - 3 u_4 = 0$$
Hence Option D is the correct answer.

Given,

sin (x + y) = cos [3(x + y)]
Using: cos θ = sin (90° - θ)
sin (x + y) = sin [90° - 3(x + y)]
sin [90° - 3(x + y)] - sin (x + y) = 0
sinC−sinD =2sin[(C−D)/2] cos[(C+D)/2]
=2sin$$\frac{(90-3(x+y)-(x+y))}{2} cos \frac{90-3(x+y)+(x+y)}{2}$$=0
=2sin(45-2(x+y)) cos (45-(x+y))=0
∴ sin 45° - 2(x + y)} = 0
45° - 2(x + y) = 0
2(x + y) = 45°
OR
Cos{45°- (x + y)} = 0
45°- (x + y)} = 90°
x + y = - 45°
2(x + y) = - 90°
Putting 2(x + y) = 45°
tan 2(x + y) = tan 45° = 1
Again, Putting 2(x + y) = - 90°, we will not get any answer among given options
Option B is the correct answer.

cot 70° = cot (90° - 20°) = tan 20°
tan 70° = tan (90° - 20°) = cot 20°
$$(1 + sec 20^{\circ} + cot 70^{\circ}) = (1 + sec 20^{\circ} + tan 20^{\circ}) $$ $$=\frac{cos 20^{\circ} + sin 20^{\circ} + 1}{cos20^{\circ}}$$

$$(1 + cosec 20^{\circ} + tan70^{\circ}) = (1 - cosec 20^{\circ} + cot20^{\circ}) $$ $$ = \frac{cos 20^{\circ} + sin 20^{\circ} - 1}{sin20^{\circ}}$$

$$(1 + sec 20^{\circ} + cot 70^{\circ}) \times (1 - cosec 20^{\circ} + tan 70^{\circ}) $$ $$=\frac{cos 20^{\circ} + sin 20^{\circ} + 1}{cos20^{\circ}} \times \frac{cos 20^{\circ} + sin 20^{\circ} - 1}{sin20^{\circ}}$$

$$= \frac{(cos 20^{\circ} + sin 20^{\circ})^2 - 1}{sin 20^{\circ}cos20^{\circ}}$$

$$= \frac{cos^2 20^{\circ} + sin^2 20^{\circ} + 2sin 20^{\circ}cos20^{\circ} - 1}{sin 20^{\circ}cos20^{\circ}}$$

$$=\frac{ 2sin 20^{\circ}cos20^{\circ} }{sin 20^{\circ}cos20^{\circ}} =2$$

Hence Option C is the correct answer.

$$4sec^2\theta+9cosec^2\theta$$
or $$4+4tan^2\theta+9+9cot^2\theta$$
or $$13+4tan^2\theta+9cot^2\theta$$
or $$ 13+4tan^2\theta+\frac{9}{tan^2\theta} $$
or $$  13-12+(2tan\theta+\frac{3}{tan\theta})^2 $$    (eq. (1) )
or now above expression to be minimum, equation $$(2tan\theta+\frac{3}{tan\theta})^2$$ should be minimum.
So applying $$A.M.\geq G.M. $$
$$\frac{(2tan\theta +\frac{3}{tan\theta})}{2} \geq \sqrt{6}$$
or $${(2tan\theta+\frac{3}{tan\theta})}=2\sqrt{6}$$ ( for value to be minimum)
After putting above value in eq.(1) , we will get least value of expression as 25.

p sin θ = √3 and p cos θ = 1
tan θ = √3
θ =60$$^o$$
p cos θ = 1
p cos 60$$^o$$ = 1
p (1/2)=1
p=2
Hence Option D is the correct answer

Given, xsinθ -ycosθ = 0
y=xsinθcosθ$\Rightarrow y=\frac{xsin\theta }{cos\mathrm{\theta <br>}}$Given the expression:
xsin$$^3$$θ + ycos$$^3$$θ = sinθcosθ
Replacing the value of y we get,
xsin$$^3$$θ+(xsinθ/cosθ)×cos$$^3$$θ=sinθcosθ
x sin3θ + x sinθcos$$^2$$θ = sinθcosθ
x sinθ × (sin2θ + cos2θ) = sinθcosθ
We know the identity: sin2θ + cos2θ = 1
x = cosθ
y=xsinθ/cosθ
y = sinθ
Hence,
x$$^2$$ + y$$^2$$ = cos$$^2$$θ + sin$$^2$$θ
= 1
Option A is the correct answer.

Euler's polyhedral formula states that the number of vertices V, faces F and edges E in a polyhedron satisfy :

V + F - E = 2

=> Ans - (B)

$$sin\frac{\pi{x}}{2}=x^{2}-2x+2$$
We know that sin can take only 3 integral values, 1,0, and -1.
$$x^{2}-2x+2$$ cannot be equal to -2. Any negative value of x will increase the value of expression beyond 4.
The value of the expression can be made equal to 1 (By substituting x = 1).
X = 1 also satisfies the LHS as sin 90 = 1.
Hence, option B is the right answer.

By Pythagoras theorem,
$$x^2 + (x-2)^2 = 20$$
$$x^2 + x^2 +4x+4 =20$$
$$2x^2+4x+4 =20$$
$$x^2+2x+2=10$$
Solving the quadratic equation we get
x=4 and x=-2
Seince x cannot be negative x=4.
AC= $$2 \sqrt{5}$$
$$Cos A = \frac{x}{2 \sqrt{5}}$$
$$Cos ^{2} A = \frac{x^{2}}{20}= \frac{16}{20}=\frac{4}{5}$$

$$Cos C =Cos (90-A) =Sin A = \frac{x-2}{2 \sqrt{5}}$$

$$Sin ^{2} A = \frac{(x-2)^{2}}{20}=\frac{4}{20}=\frac{1}{5}$$

$$(cos^2 A - cos^2 C) = Cos ^{2} A - Sin ^{2} A = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}$$
Hence Option B is the correct answer.

$$tan(x + y) = \frac{tan x + tan y}{1- tan x tan y}$$

$$tan(x - y) = \frac{tan x - tan y}{1+ tan x tan y}$$

$$tan(x+y)tan(x-y) =1$$

$$ \frac{tan x + tan y}{1- tan x tan y} \times \frac{tan x - tan y}{1+ tan x tan y} =1 $$

$$ \frac{tan^{2} x - tan^{2} y}{1 - tan^{2} x tan ^{2} y} =1$$

$$ tan^{2} x - tan^{2} y =1 - tan^{2} x tan ^{2} y$$

$$ tan^{2} x + tan^{2} x tan^{2} y =1 + tan ^{2} y$$

$$ tan^{2} x (1 + tan^{2} y) =1 + tan ^{2} y$$

$$ tan^{2} x =1$$

$$ tan x =1$$

Option B is the correct answer.

Expression : $$cosec2A + cot2A$$

= $$\frac{1}{sin2A}+\frac{cos2A}{sin2A}$$

= $$\frac{1+cos2A}{sin2A}$$

= $$\frac{1+(2cos^2A-1)}{2sinAcosA}$$

= $$\frac{2cos^2A}{2sinAcosA}$$

= $$\frac{cosA}{sinA}=cotA$$

=> Ans - (C)

Given : A = 30°, B = 60° and C = 135°

To find : $$sin^3A + cos^3B + tan^3C - 3sin A cos B tan C$$

= $$[sin(30^\circ)]^3+[cos(60^\circ)]^3+[tan(135^\circ)]^3-3[sin(30^\circ)][cos(60^\circ)][tan(90^\circ)]$$

= $$(\frac{1}{2})^3+(\frac{1}{2})^3+(-1)^3-3(\frac{1}{2})(\frac{1}{2})(-1)$$

= $$\frac{1}{8}+\frac{1}{8}-1+\frac{3}{4}$$

= $$\frac{1}{4}-\frac{1}{4}=0$$

=> Ans - (A)

Expression : $$tan^2θ + cot^2θ + sin^2θ + cos^2θ + sec^2θ + cosec^2θ$$

Using, $$(sec^2\theta-tan^2\theta=1)$$ and $$(cosec^2\theta-cot^2\theta=1)$$

= $$(sin^2\theta+cos^2\theta)+tan^2\theta+cot^2\theta+(1+tan^2\theta)+(1+cot^2\theta)$$

= $$1+1+1+2tan^2\theta+2cot^2\theta$$

= $$3+2(tan^2\theta+cot^2\theta)$$

Minimum value of $$(tan^2\theta+cot^2\theta)=2$$

= $$3+2(2)=3+4=7$$

=> Ans - (D)

As we can see, the least value of the expression is 5/4 = 1.25>1.
Cos value cannot be greater than 1. If it is greater than 1, it means that adjacent side > hypotenuse. Hence, hypotenuse will not remain the hypotenuse anymore. Hence, option D is the right answer.

We need to find the value of $$1+\frac{1}{\cot^{2}63^{\circ}}-\sec^{2}27^{\circ}+\frac{1}{\sin^{2}63^{\circ}}-cosec^{2}27^{\circ}$$

we know that ,

$$\frac{1}{cot^2 \theta}$$ = $$tan^2 \theta$$.............(1)

$$\frac{1}{sin^2 \theta}$$ = $$cosec^2 \theta$$...........(2)

and 1 + $$tan^2 \theta$$ = $$sec^2 \theta$$.................(3)

Using equations 1 ,2 and 3

= $$1+\frac{1}{\cot^{2}63^{\circ}}-\sec^{2}27^{\circ}+\frac{1}{\sin^{2}63^{\circ}}-cosec^{2}27^{\circ}$$

= 1 + $$tan^2 63$$ - $$sec^2 27$$ + $$cosec^2 63$$ - $$cosec^2 27$$

= $$sec^2 63$$ - $$sec^2 27$$ + $$cosec^2 63$$ - $$cosec^2 27$$

= $$sec^2 (90-27)$$ - $$sec^2 27$$ + $$cosec^2 (90-27)$$ - $$cosec^2 27$$

=$$cosec^2 27$$ - $$sec^2 27$$ + $$sec^2 27$$ - $$cosec^2 27$$

= 0

From the diagram,
Height = AD = 50√3 m
∠BAN = 30°
∠CAM = 60°
∴∠BAD = 90° - 30° = 60°
∴∠CAD = 90° - 60° = 30°
From ΔABD,
tan∠BAD = Perpendicular/ Base
tan60° = BD/AD
√3 = BD/(50√3)
BD = 50 × 3 = 150 m
From ΔACD,
tan∠CAD = Perpendicular/ Base
tan30° = CD/AD
1/√3 = CD/(50√3)
CD = 50 m
∴ Distance travelled by the bird
= BC = BD + CD = 150 m + 50 m = 200 m = 0.200 km
Time taken to cover this distance = 2 minutes = 2/60 hr = 1/30 hr
∴ Speed
= Distance travelled/ Time required
=0.200 $$\times$$30km/hr$\mathrm{<br>}$= 0.200 × 30 km/hr
= 6 km/hr
Option D is the correct answer

Sin(2a) = 2sin(a/2)cos(a/2)
Cos(2a) = 2cos$$^2$$a - 1 = 1 - 2sin$$^2$$a
Sin$$^2$$a + cos$$^2$$a = 1
cosec$$^2$$a - cot$$^2$$a = 1
Given,
tanθ - cotθ = a
(sinθ/cosθ)−(cosθ/sinθ)=a
(sin$$^2$$θ−cos$$^2$$θ)/cosθsinθ=a
=−2cos2θ/sin2θ=a
a = -2cot2θ
Also given, cosθ - sinθ = b
Squaring both sides and using (a - b)$$^2$$ = a$$^2$$ + b$$^2$$ - 2ab, we get,
Cos$$^2$$θ + sin$$^2$$θ - 2cosθsinθ = b$$^2$$
1 - sin2θ = b$$^2$$
We have to find the value of
(a$$^2$$ + 4) (b$$^2$$ - 1)$$^2$$
(4cot$$^2$$2θ + 4)(1 - sin2θ - 1)2
4(cosec$$^2$$2θ)(-sin2θ)$$^2$$
= 4
Option A is the correct answer.

Cos(a - b) = cosa cosb + sina sinb

(a$$^2$$ - b$$^2$$) sinθ + 2ab cosθ = a$$^2$$ + b$$^2$$
$$\frac{a^2 - b^2}{a^2 + b^2} sin \theta + \frac{2ab}{a^2 + b^2} cos \theta = a^2 + b^2$$
Let $$\frac{(a^2 - b^2)}{(a^2 - b^2)}$$ = sin A,
then $$ \frac{2ab}{a^2 + b^2} $$$\frac{{\mathrm{}}^{}}{}$= cosA
sinAsinθ + cosAcosθ = 1
cos (A - θ) = 1
A - θ = 0° (as, cos 0° = 1)
θ = A
∴ tanθ = tanA
tanθ = sinA/cosA
tanθ= $$\frac{\frac{(a^2 - b^2)}{(a^2 - b^2)}}{\frac{2ab}{a^2 + b^2}}$$
tanθ= $$\frac{(a^2 - b^2)}{2ab}$$
Option C is the correct answer.

$$2y=tan\theta$$
$$x=2ycosec\theta$$
Hence value of $$x^2 - 4y^2 $$ = $$4y^2(cosec^2\theta - 1)$$
or $$tan^2\theta cot^2\theta$$ = 1

Expression : $$x cosθ - sinθ = 1$$

=> $$x = \frac{1}{cos\theta} + \frac{sin\theta}{cos\theta}$$

=> $$x = sec\theta + tan\theta$$ --------------Eqn(1)

$$\because$$ $$sec^2\theta - tan^2\theta = 1$$

=> $$(sec\theta + tan\theta)(sec\theta - tan\theta) = 1$$

=> $$(sec\theta - tan\theta) = \frac{1}{x}$$ --------------Eqn(2)

Adding eqns(1)&(2)

=> $$2sec\theta = x + \frac{1}{x} = \frac{x^2 + 1}{x}$$

=> $$sec\theta = \frac{x^2 + 1}{2x}$$

Subtracting eqn(2) from (1)

=> $$2tan\theta = x - \frac{1}{x} = \frac{x^2 - 1}{x}$$

=> $$tan\theta = \frac{x^2 - 1}{2x}$$

We know that, $$sin\theta = \frac{tan\theta}{sec\theta}$$

=> $$sin\theta = \frac{x^2 - 1}{2x} * \frac{2x}{x^2 + 1}$$

=> $$sin\theta = \frac{x^2 - 1}{x^2 + 1}$$

To find : $$x^2 - (1 + x^2) sinθ $$

= $$x^2 - (1 + x^2) * \frac{x^2 - 1}{x^2 + 1}$$

= $$x^2 - (x^2 - 1)$$

= 1

Taking $$cos\theta$$ outside in numerator and in denominator and making $$tan\theta$$
hence eq will be  $$(\frac{5tan\theta - 3}{5tan\theta + 3})$$
As it is given that $$5tan\theta$$ = 4
after putting values and solving we will get the equation reduced to 1/7.

$$x=cosec\theta - sin\theta=\frac{cos^2\theta}{sin\theta}=cot\theta cos\theta$$
Similarly $$y=tan\theta sin\theta$$
$$xy=sin\theta cos\theta$$
$$x^2+y^2+3=(sec^2\theta +cosec^2\theta )$$
Now putting above values in given equation, and after solving it will be reduced to 1

Given equation can be written as $$(cos^4\theta + cos^2\theta)^3 -1$$
as $$sin\theta + sin^2\theta = 1$$
or $$sin\theta = cos^2\theta$$
putting above value in given equation it will be
$$(sin^2\theta + sin\theta)^3 -1 = 0$$

for  asin θ + bcos θ + c,

maximum value = $$c+\sqrt{a^{2}+b^{2}}$$

minimum value =  $$c-\sqrt{a^{2}+b^{2}}$$

for  sin θ + cos θ , a = 1, b = 1, c = 0

maximum value = $$c+\sqrt{a^{2}+b^{2}}=0+\sqrt{1^{2}+1^{2}}=\sqrt{2}$$

so the answer is option B.

Expression : $$(\frac{2}{\sqrt{3}}+tan45^\circ)$$

= $$\frac{2}{\sqrt3}+1$$

= $$\frac{(2+\sqrt{3})}{\sqrt{3}}$$

=> Ans - (B)

sin²$$\alpha$$ +cos²$$\alpha$$ =1 (identity)
cos²$$\alpha$$ = 1-sin²$$\alpha$$
2sin$$\alpha$$ +15cos²$$\alpha$$ =7
put 1-sin²$$\alpha$$ instead of cos²$$\alpha$$
2sin$$\alpha$$ +15(1-sin²$$\alpha$$) =7
-15sin²$$\alpha$$ +2sin$$\alpha$$ +8=0
Let sin$$\alpha$$ = x
-15x² +2x +8=0
Solving for x we get,
x= 4/5 and x = -2/3
x = 4/5 is the real solution
sin $$\alpha$$= 4/5
sin² $$\alpha$$= 16/25
sin²$$\alpha$$+cos²$$\alpha$$=1 = sin²$$\alpha$$ = 1-cos²$$\alpha$$
1-cos²$$\alpha$$ =16/25 = cos²$$\alpha$$ =9/25 = cos$$\alpha$$ =3/5
cot $$\alpha$$ = cos $$\alpha$$ / sin $$\alpha$$ = (3/5) / (4/5) = 3/4
Option A is the correct answer.

we need to find value of $$\frac{sin 43^{\circ}}{cos 47^{\circ}}+\frac{cos 19^{\circ}}{sin 71^{\circ}}-8cos^{2}60^{\circ}$$

$$\frac{sin (90-47)}{cos 47}+\frac{cos (90-17)}{sin 71}-8cos^{2}60$$

$$\frac{cos47)}{cos 47}+\frac{sin17)}{sin 71}-8( \frac{1}{2})^2$$

= 0

Expression : $$\frac{cos^2 45}{sin^2 60}+\frac{cos^2 60}{sin^2 45}-\frac{tan^230}{cot^245}-\frac{sin^230}{cot^230}$$

= $$\frac{(\frac{1}{\sqrt{2}})^2}{(\frac{\sqrt{3}}{2})^2} + \frac{(\frac{1}{2})^2}{(\frac{1}{\sqrt{2}})^2} - \frac{(\frac{1}{\sqrt{3}})^2}{(1)^2} - \frac{(\frac{1}{2})^2}{(\sqrt{3})^2}$$

= $$(\frac{1}{2} \times \frac{4}{3}) + (\frac{1}{4} \times 2) - (\frac{1}{3} \times 1) - (\frac{1}{4 \times 3})$$

= $$\frac{2}{3} + \frac{1}{2} - \frac{1}{3} - \frac{1}{12}$$

= $$\frac{9}{12} = \frac{3}{4}$$

Expression : $$2sin^{2}$$ θ + $$3cos^{2}$$ θ

We know that, $$sin^2 \theta = 1 - cos^2 \theta$$

=> $$2 (1- cos^2 \theta) + 3cos^2 \theta$$

= $$cos^2 \theta + 2$$

Using the formula, $$cos^2 \theta = \frac{cos 2\theta + 1}{2}$$

=> $$\frac{cos 2\theta + 1}{2} + 2$$

= $$\frac{cos 2\theta}{2} + \frac{5}{2}$$

$$\because$$ minimum value of $$cos 2\theta = -1$$

=> min value = $$\frac{5}{2} - \frac{1}{2} = 2$$

Expression : tan 4° tan 43° tan 47 tan 86°

$$\because$$ $$tan(90-\theta) = cot\theta$$

=> $$tan 4^{\circ} = tan(90^{\circ}-86^{\circ}) = cot 86^{\circ}$$

Similarly, $$tan 43^{\circ} = cot 47^{\circ}$$

=> $$(cot 86^{\circ} \times tan 86^{\circ}) * (tan 47^{\circ} \times cot 47^{\circ})$$

Using, $$tan\theta cot\theta$$ = 1

=> 1*1 = 1

Expression : $$sin\theta + sin^2\theta = 1$$

=> $$sin\theta = 1 - sin^2\theta$$

=> $$sin\theta = cos^2\theta$$

To find : $$cos^2\theta + cos^4\theta$$

= $$cos^2\theta + (cos^2\theta)^2$$

= $$cos^2\theta + sin^2\theta$$

= 1

$$\frac{\cos\alpha}{\sin\beta}=n$$

=> $$cos\alpha = nsin\beta$$

and $$\frac{\cos\alpha}{\cos\beta}=m$$

=> $$cos\alpha = mcos\beta$$

Comparing above equations, we get :

=> $$nsin\beta = mcos\beta$$

Squaring both sides :

=> $$n^2sin^2\beta = m^2cos^2\beta$$

=> $$n^2(1-cos^2\beta) = m^2cos^2\beta$$

=> $$n^2 = (n^2+m^2)cos^2\beta$$

=> $$cos^2\beta = \frac{n^2}{m^2+n^2}$$

Expression : $$tan\theta + cot\theta = 5$$

Squaring both sides, we get :

=> $$tan^2\theta + cot^2\theta + 2tan\theta cot\theta = 25$$

We know that, $$tan\theta cot\theta = 1$$

=> $$tan^2\theta + cot^2\theta = 25-2 = 23$$

Height of tower = AB

In $$\triangle$$ABC

=> $$tan\theta = \frac{AB}{BC}$$

=> $$tan30^{\circ} = \frac{AB}{100}$$

=> $$\frac{1}{\sqrt{3}} = \frac{AB}{100}$$

=> $$AB = \frac{100}{\sqrt{3}}$$

If secθ - cosecθ = 0 Then θ = 45$$^o$$
sec 45$$^o$$ =cosec 45$$^o$$ = $$\sqrt{2}$$
secθ + cosecθ = $$2 \times \sqrt{2}$$
Hence Option D is the correct answer.

Expression : $$sin A cos A (tan A - cot A)$$

= $$sin A cos A (\frac{sin A}{cos A} - \frac{cos A}{sin A})$$

= $$sin A cos A (\frac{sin^2 A - cos^2 A}{sin A cos A})$$

= $$sin^2 A - cos^2 A$$

= $$sin^2 A - (1 - sin^2 A)$$

= $$2sin^2 A - 1$$

=> $$sec\theta + tan\theta = p$$ --------------Eqn(1)

$$\because$$ $$sec^2\theta - tan^2\theta = 1$$

=> $$(sec\theta + tan\theta)(sec\theta - tan\theta) = 1$$

=> $$(sec\theta - tan\theta) = \frac{1}{p}$$ --------------Eqn(2)

Adding eqns(1)&(2)

=> $$2sec\theta = p + \frac{1}{p} = \frac{p^2 + 1}{p}$$

=> $$sec\theta = \frac{1}{2}[p + \frac{1}{p}]$$

given that tan $$\theta$$ + cot $$\theta$$ = 2

$$\frac{sin \theta}{cos \theta}$$ + $$\frac{cos \theta}{sin \theta}$$ = 2

$$sin^2 \theta + cos^2 \theta$$ = 2 $$sin \theta cos \theta $$

$$sin 2\theta$$ = 1

it implies , $$2\theta$$ = 90

$$\theta$$ = 45

Expression : $$sin^{2}$$ 22° + $$sin^{2}$$ 68° + $$cot^{2}$$ 30°

We know that $$sin(90^{\circ}-\theta) = cos\theta$$

=> $$sin 22^{\circ} = sin (90^{\circ}-68^{\circ}) = cos 68^{\circ}$$

=> $$cos^2 68^{\circ} + sin^2 68^{\circ} + (\sqrt{3})^2$$

=> 1 + 3 = 4

tan(90-θ)=cotθ
tan(4θ-50) = cot(90 - (4θ-50))
=cot(140-4θ)
Given tan (4θ - 50°) = cot(50° - θ)
tan (4θ - 50°) = cot(140° - 4θ)
cot(50° - θ) = cot(140° - 4θ)
50° - θ = 140° - 4θ
3θ = 90°
θ = 30°
Option D is the correct answer.

We know that cosecα = 1/ sinα , hence applying A.M ≥ G.M logic, we get
A.M of given equation = (4 cosec$$2$$α + 9 sin$$^2$$α) / 2 …. (1)
G.M of given equation = √ (4 cosec$$^2$$α . 9 sin$$^2$$α )
= $$\sqrt{4 * 9}$$
= $$\sqrt{36}$$ = $$6$$ …. (2)
Now, we know that A.M ≥ G. M
From equations (1) and (2) above we get,
=>(4 cosec$$^2$$α + 9 sin$$^2$$α) / 2 ≥ 6
Multiplying both sides by 2
(4 cosec$$^2$$α + 9 sin$$^2$$α) ≥ 12
The minimum value will be 12.
Option D is the correct answer.

$$ \frac{sin^2 A}{1 + cos A} - \frac{sin A}{1 - cos A} = \frac{sin^2 A(1 - cos A)+sin A(1+cos A)}{1^{2} - cos^{2} A} =\frac{sin^2 A(1 - cos A)+sin A(1+cos A)}{ sin^{2} A} $$

$$\frac{sin^2 A(1 - cos A)+sin A(1+cos A)}{ sin^{2} A} = (1-cos A) + \frac{(1+cos A)}{sin A}$$

$$ \frac{sin^2 A}{1 + cos A} - \frac{sin A}{1 - cos A}= (1-cos A) + \frac{(1+cos A)}{sin A}$$

$$1 - \frac{sin^2 A}{1 + cos A} + \frac{1 + cos A}{sin A} - \frac{sin A}{1 - cos A}$$

$$= 1 -[(1-cos A) + \frac {1+cos A}{sin A} ] + \frac{(1+cos A)}{sin A}$$

$$= 1 -(1-cos A) - \frac {1+cos A}{sin A} + \frac{(1+cos A)}{sin A} = cos A$$
Hence Option A is the correct answer.

Expression : tan1°tan2°tan3° ……………tan88°tan89°

$$\because$$ $$tan(90^{\circ}-\theta) = cot\theta$$

=> tan 89° = tan(90°-1) = cot 1°

Similarly, tan 88° = cot 2° and so on till tan 46° = cot 44°

=> (tan1°tan2°tan3°.......tan45°......cot3°cot2°cot1°)

Using, $$tan\theta cot\theta$$ = 1 and $$tan45^{\circ}$$ = 1

=> 1*1 = 1

Expression : $$\tan^2\theta+\frac{1}{\tan^2\theta}=2$$

=> $$(tan\theta + \frac{1}{tan\theta})^2 - 2 = 2$$

=> $$(tan\theta + \frac{1}{tan\theta})^2 = 4$$

=> $$tan\theta + \frac{1}{tan\theta} = 2$$

[It can't be -2 as $$\theta$$ is in 1st quadrant, and $$tan\theta$$ is positive in 1st quadrant.]

=> $$tan^2\theta + 1 = 2tan\theta$$

=> $$(tan\theta - 1)^2 = 0$$

=> $$tan\theta = 1$$

=> $$\theta = 45^{\circ}$$

Height of person = CD = 6 ft

Height of tree = AB = $$\frac{26}{3}$$ ft

Distance between them = BD = $$\frac{8}{\sqrt{3}}$$ ft

To find : $$\angle$$ACE = $$\theta$$ = ?

Solution : AE = AB - BE = $$\frac{26}{3}$$ - 6

=> AE = $$\frac{8}{3} ft$$

and BD = CE = $$\frac{8}{\sqrt{3}}$$ ft

Now, in $$\triangle$$AEC

=> $$tan\theta$$ = $$\frac{AE}{CE}$$

=> $$tan\theta$$ = $$\frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}}$$

=> $$tan\theta$$ = $$\frac{1}{\sqrt{3}}$$

=> $$\theta$$ = 30°

Height of kite from ground = AB = 75 m

$$\angle$$ACB = $$\theta$$

We know that $$cot\theta = \frac{8}{15}$$

=> $$\frac{BC}{AB} = \frac{8}{15}$$

=> $$BC = \frac{8*75}{15} = 40 m$$

Now, length of string AC = $$\sqrt{(AB)^2 + (BC)^2}$$

=> AC = $$\sqrt{75^2 + 40^2}$$

= $$\sqrt{5625+1600} = \sqrt{7225}$$

=> AC = 85 m

2 sin α + 15 cos$$^2$$ α = 7
Using: cos2 α = 1 - sin$$^2$$ α
2 sin α + 15 (1 - sin$$^2$$ α) = 7
2 sin α + 15 - 15 sin$$^2$$α = 7
15 sin$$^2$$ α - 2 sin α - 8 = 0
15 sin$$^2$$ α - 12 sin α + 10 sin α - 8 = 0
3 sin α (5 sin α - 4) + 2(5 sin α - 4) = 0
(5 sin α - 4)(3 sin α + 2) = 0
5 sin α - 4 = 0 or 3 sin α + 2 = 0
sin α = 4/5 or sin α = - 2/3
∵ 0 ≤ α ≤ π/2
∴ sin α = 4/5
cosec α = 5/4
cosec$$^2$$α = 25/16
1 + cot$$^2$$α = cosec$$^2$$α
1 + cot$$^2$$α = 25/16
cot$$^2$$α = (25/16) - 1 = 9/16
cot α = 3/4[∵ 0 ≤ α ≤ π/2]
Option C is the correct answer.

If sin2$$\theta$$ = 1/2
2$$\theta$$ = 30$$^{o}$$
$$\theta$$=15$$^{o}$$
cos(75$$^{o}$$- $$\theta$$) = cos(75$$^{o}$$ -15$$^{o}$$)=cos 60$$^{o}$$ = 1/2
Hence Option B is the correct answer.

cosec A + cot A =3
cosec²A-cot²A=1;
cosec A-cot A = 1/(cosec A+cot A)
So cosec A - cot A =1/3
2 cosec A=10/3 or cosec A=5/3.
Hence sin A=3/5
cos²a=1-sin²a
So ,cos²a= 1 - (9/25) = 16/25
cos A=4/5.
Option A is the correct answer.

Let the unknown angle of depression be x.Since the angles of depression are complementary, the other angle is (90-x)
$$\angle ROQ = x$$ since $$\triangle$$ ROQ is a right angled triangle.
$$h = 10 \sqrt{3}$$
Let RQ = y metres
tan x = $$\frac{OR}{RP} = \frac{RQ}{OR}$$
$$\frac{OR}{RP}=\frac{h}{y+20}$$
$$\frac{RQ}{OR} = \frac{y}{h}$$
$$\frac{h}{y+20} = \frac{y}{h}$$
$$\frac{10\sqrt{3}}{y+20} = \frac{y}{10\sqrt{3}}$$
$$y(y+20)= 300$$
Solving for y we get y = 10 m.
RP = 20 +10 m = 30m is the distance of P from the building.
Option C is the correct answer.