# 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**

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**Joshi sir comment**

First replace cotB by cotA.

because it is wrongly printed by you.

answer by NIKHIL is correct

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?????

**Asked By: AMBUJ SINGLA**

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**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

**Asked By: AMBUJ SINGLA**

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**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 ;

**Asked By: AMIT DAS**

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**Joshi sir comment**

by 1st equation sec y = √(1+cos^{2}x) so cos y = 1/ √(1+ cos^{2}x)

by 3rd equation sec z = cot x so tan z = √(cot^{2}x - 1)

on putting these 2 values in 2nd equation we get

1/ √(1+ cos^{2}x) = √(cot^{2}x - 1)

so sin^{2}x = (1+cos^{2}x)(cos^{2}x-sin^{2}x)

or sin^{2}x = (2-sin^{2}x)(1-2sin^{2}x)

or sin^{2}x = 2-4sin^{2}x-sin^{2}x+2sin^{4}x

or 2sin^{4}x-6sin^{2}x+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..

**Asked By: AMIT DAS**

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**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 sec^{2}x = dy/dx

on comparing with 0 we get d tanx = a secx or sinx = a/d so secx = d/√(d^{2}-a^{2}) and tanx = a/√(d^{2}-a^{2})

so minimum of given expression = [d^{2}/ √(d^{2}-a^{2})] - [a^{2}/√(d^{2}-a^{2})] = √(d^{2}-a^{2})

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).

**Asked By: AMIT DAS**

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**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|.

**Asked By: AMIT DAS**

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**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 *z*^{15} = 1 and Iarg zI< π/2 is

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

let z = r(cosθ+i sinθ) and it is given that z^{15} = 1 => r^{15}(cosθ+i sinθ)^{15 }= 1

or r^{15}(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 .

**Asked By: HIMANSHU MITTAL**

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**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).

**Asked By: HIMANSHU MITTAL**

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**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