# 30 - Trigonometry Questions Answers

if secA-tanA=3

then find the value of (5cosA+4cotB)??

jus a hint

Asked By: AMBUJ SINGLA Solved By: NIKHIL VARSHNEY
Joshi sir comment

First replace cotB by cotA.

because it is wrongly printed by you.

Which of the following angles are co-terminal angles: 480  ,   -240  or 600

how can we get it which is co-terminal and what does co-terminal means?????

Joshi sir comment

Coterminal angles are angles in standard position (angles with the initial side on the positive x-axis) that have a common terminal side.  For example 30°, –330° and 390° are all coterminal.

To find a positive and a negative angle coterminal with a given angle, you can add and subtract 360° if the angle is measured in degrees or 2π if the angle is measured in radians.

480 and -240 are co terminal angles because 480-360 = 120 and -240+360 = 120

is there any need to know or learn the proof pf formulas of trigo

Joshi sir comment

No, For making a good command in trigonometry, only implementation is a need. But for your knowledge you can proof these results

if x,y,z are acute and cos x= tan y , cos y = tanz , cos z = tan x, then the value of sin x is ;

Joshi sir comment

by 1st equation  sec y = √(1+cos2x)  so cos y = 1/ √(1+ cos2x)

by 3rd equation sec z = cot x  so  tan z = √(cot2x - 1)

on putting these 2 values in 2nd equation we get

1/ √(1+ cos2x) = √(cot2x - 1)

so sin2x = (1+cos2x)(cos2x-sin2x)

or sin2x = (2-sin2x)(1-2sin2x)

or sin2x = 2-4sin2x-sin2x+2sin4x

or 2sin4x-6sin2x+2 = 0

now solve it for sinx

If     a sinx + b cos(x+θ) + b cos(x-θ) = d, then the minimum value of |cos θ| is equal to :

sir i have expanded cos(x+θ) & cos(x-θ) and after that i got the equation: (d sec x - a tanx)/2b = cos θ. now , can i write  minimum value of (d sec x -a tanx) = (d²-a²)^(1/2).. as all the options contain (d²-a²)^(1/2).......or do tell the other hint to solve it..

Joshi sir comment

Dear amit your starting steps are correct

now let us consider that   d secx - a tanx  = y

on differentiating we will get d secx tanx - a sec2x = dy/dx

on comparing with 0 we get d tanx = a secx       or               sinx = a/d               so         secx = d/√(d2-a2)           and   tanx = a/√(d2-a2)

so minimum of given expression = [d2/ √(d2-a2)] - [a2/√(d2-a2)]  =        √(d2-a2)

The distance between two parallel lines is 1.A point 'A' is chosen to lie between the  lines at a distance 'd' from the first line. Triangle ABC is an eqilateral triangle with 'B" on one line and 'C' on the other line.Then AB is equal to(in terms of d).

Joshi sir comment

In triangle BCX

BC = seca

so AB = seca

now in triangle ABY

cos(60+a) = d/seca

so 1/2 cosa - √3/2 sina = d cosa

or tana = (1-2d)/√3

now calculate seca

let z be a complex number satisfying |z-3|=|z-4i|, then find the least possible value of 10|z|.

Joshi sir comment

In |z-3|=|z-4i|, z represents all points lying on the perpendicular bisector of the line joining 3 and 4i in x-y plane. In this perpendicular bisector, if we draw a perpendicular then it will be the minimum magnitude of z. Construct this as a diagram we will get

sinθ = |z| / (7/8)

or |z| = (7/8)(4/5) = 7/10

The number of roots of equations z15 = 1 and Iarg zI< π/2 is

Joshi sir comment

let z = r(cosθ+i sinθ) and it is given that z15 = 1  =>  r15(cosθ+i sinθ)15 = 1

or r15(cos15θ+i sin15θ) = 1+0i                   (Here we use Demoivre's theorem)

on comparison we get sin 15θ = 0 => 15θ = nπ          here n is an integer

for n = -7 to +7 we will get 15 answers for which θ will lie between -π/2 to +π/2

Find the number of solutions of the equation 2cos²(x/2)sin²(x) = x² + 1/x² , 0 < x ≤ π/2 .

Joshi sir comment

max of sin and cos are 1 so max of left is 2

but x² + 1/x² ≥ 2x(1/x) = 2

so only solution will be obtained for 2cos²(x/2)sin²(x) = 2

so (1+cosx)(1-cos²x) = 2 which is impossible,so i think no solution

Find the value of 16cos(2π/15)cos(4π/15)cos(8π/15)cos(14π/5).

Joshi sir comment

I think the last term would be cos(14π/15)

We know that cos(14π/15) = -cos(π/15)

so     16cos(2π/15)cos(4π/15)cos(8π/15)cos(14π/15)

= -16cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

= -[8/sin(π/15)]2sin(π/15)cos(π/15)cos(2π/15)cos(4π/15)cos(8π/15)

=  -[8/sin(π/15)]sin(2π/15)cos(2π/15)cos(4π/15)cos(8π/15)

take the similar steps