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1

Answer:

As below.

Explanation:

Standard form of Secant function is #color(violet)(y = A sec (Bx - C) + D#

#"Given " y = - sec( 2x)#

#A = -1, B = 2, C = 0, D = 0#

#Amplitude = "None for secant function"#

#"Period " = (2pi) / |B| = (2pi)/ 2 = pi#

#"Phase Shift " = -C / B = 0#

#"Vertical Shift " = D = 0#

graph{-sec (2x) [-10, 10, -5, 5]}

2

Answer:

#:. alpha=2kpi+-arccos[{-(sqrt2-1)+-root(4)2}/3], k in ZZ#.

Explanation:

Replacing #sin^2alpha# in the right member by #(1-cos^2alpha)#, we have,

#(2sqrt2-2)cosalpha=3-3cos^2alpha+sqrt2-4#.

#:. 3cos^2alpha+2(sqrt2-1)cosalpha-(sqrt2-1)=0#.

This is a quadr. eqn. in #cosalpha#, and, its discriminant

#Delta=(2sqrt2-2)^2+4*3*(sqrt2-1)#,

#=(8-8sqrt2+4)+12sqrt2-12#,

# :. Delta=4sqrt2#.

Therefore, using the quadr. formula, we have,

#cosalpha={-2(sqrt2-1)+-sqrt(4sqrt2)}/(2*3)#,

#={-2(sqrt2-1)+-2root(4)2}/(2*3)#,

#:. cosalpha={-(sqrt2-1)+-root(4)2}/3#.

#:. alpha=2kpi+-arccos[{-(sqrt2-1)+-root(4)2}/3], k in ZZ#.

1

Answer:

Kindly see a Proof in the Explanation.

Explanation:

#(sectheta-1)/(sectheta+1)#,

#=(1/costheta-1)-:(1/costheta+1)#,

#=(1-costheta)/costheta-:(1+costheta)/costheta#,

#=(1-costheta)/(1+costheta)#,

#=(1-costheta)/(1+costheta)xx(1+costheta)/(1+costheta)#,

#=(1-cos^2theta)/(1+costheta)^2#,

#=sin^2theta/(1+costheta)^2#, as desired!

1

Answer:

#x=120^circ or240^circ#

Explanation:

Here,

#2cosx+1=0 , where, 0^circ <= x <= 360^circ#

#=>2cosx=-1#

#=>cosx=-1/2 < 0=>II^(nd)Quadrant or III^(rd)Quadrant#

#(i)90^circ < x < 180^circ=>x=120^circto(II^(nd)Quadrant)#

#(ii)180^circ < x < 270^circ=>x=240^circto(III^(rd)Quadrant)#

Hence,

#x=120^circ or240^circ#

1

Answer:

We know that,
#color(blue)(cosC+cosD=2cos((C+D)/2)cos((C-D)/2)to(1)#

Explanation:

We have to prove ,

#sin((5pi)/18)+cos((4pi)/9)=cos(pi/9)#

We take,

#LHS=sin((5pi)/18)+cos((4pi)/9)#

#color(white)(LHS)=cos(pi/2-(5pi)/18)+cos((4pi)/9)#

#color(white)(LHS)=cos((9pi-5pi)/18)+cos((4pi)/9)#

#color(white)(LHS)=color(blue)(cos((2pi)/9)+cos((4pi)/9)...toApply(1)#

#color(white)(LHS)=color(blue)(2cos(((2pi)/9+(4pi)/9)/2)cos(((2pi)/9-(4pi)/9)/2)#

#color(white)(LHS)=2cos((6pi)/18)cos((-2pi)/18)#

#color(white)(LHS)=2cos(pi/3)cos(-pi/9)#

#color(white)(LHS)=2(1/2)cos(pi/9)...to[becausecos(- theta)=costheta ]#

#color(white)(LHS)=cos(pi/9)#

#LHS=RHS#

2

Answer:

#"Screen height " = 15.2 " meters"#

Explanation:

Anjali G

Using the variable definitions of #a# and #b# as shown in the diagram above, we can use trigonometry to solve for the height of the screen, #a+b#.

#tan(13^@) = a/11#

#a = 11tan(13^@)#

#tan(49^@) = b/11#

#b = 11tan(49^@)#

Therefore, the height of the screen #a+b# is:

#a + b = 11tan(13^@)+11tan(49^@)#

#a+b = 2.5396 + 12.6541#

#a+b = 15.1936 " meters"#

1

Answer:

#color(blue)((2 + i) / (4 + i9) = 0.175 - i 0.144#

Explanation:

To divide #(2 + i) / (4 + i9)# using trigonometric form.

#z_1 = (2 + i), z_2 = (4+ i9)#

#r_1 = sqrt(2^2 + 1^2) = sqrt5#

#r_2 = sqrt(9^2 + 4^2) = sqrt97#

#theta_1 = arctan (1/2) = 26.57^@#

#Theta_2 = arctan(9/4) = 66.06^@#

#r_1 / r_2 = sqrt 5 * / sqrt 97 ~~ 0.227#

#z_1 / z_2 = (r_1 / r_2) * (cos (theta_1 - theta_2) + i sin (theta_1 - theta_2))#

#z_1 / z_2 = (0.227) * (cos (26.57 - 66.06 ) + i sin (26.57 - 66.06 ))#

#z_1 / z_2 = 0.227 * (cos (-39.49) + i sin (-39.49)) = 0.227 (0.772 - i 0.636)#

#color(blue)((2 + i) / (4 + i9) = 0.175 - i 0.144#

1

Answer:

#color(maroon)(=> -5sqrt(3/2) + i 5sqrt(3/2)# or #-5sqrt(3/2)(1-i)#

Explanation:

#Z_1 * Z_2 = r_1*r_2 (cos (theta_1+ theta_2) + i sin (theta_1 + theta_2))#

#Z_1 = 5 (cos ((3pi)/4) + i sin ((3pi)/4)#

#Z_2 = sqrt3 (cos (pi/2) + i sin (pi/2))#

#r_1 =5, r_2 = sqrt3, theta_1 = (3pi)/4, theta_2 = pi/2#

#Z_1 * Z_2 = 5 sqrt3 ((cos ((3pi)/4 + pi/2) + i sin ((3pi)/4 + pi/2))#

#=> 5 sqrt3 * (-(1/sqrt2) + i (1/sqrt2))#

#color(maroon)(=> -5sqrt(3/2) + i 5sqrt(3/2)# or #-5sqrt(3/2)(1-i)#

1

Answer:

#tan(B/4)=(sqrt13-3)/2#

Explanation:

Here,

#a=5 , b=12 and c=13#

#=>BC=5, CA=12 ,AB=13#

#=>bar(AB) # , is the hypotenuse of #triangleABC#

So, #mangleC=90^circ=>angle A and angle B# are acute angles.

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Now, #color(red)(cosB=("side adjacent to B")/("hypotenus")=5/13#

Using half angle formula :

#cos(B/2)=(1+cosB)/2=(1+5/13)/2=9/13=(3/sqrt13)^2#

#=>color(red)(cos(B/2)=+3/sqrt13...to[because mangleB < 90^circ]#

Also ,we know the half angle formula:

#color(blue)(tan^2(theta/2)=(1-costheta)/(1+costheta)#

Let us take, #theta=(B/2)#

#:.color(blue)(tan^2((B/2)/2)=(1-cos(B/2))/(1+cos(B/2))#

#=>tan^2(B/4)=(1-3/sqrt13)/(1+3/sqrt13)=(sqrt13-3)/(sqrt13+3)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((1)#

#=>tan^2(B/4)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((sqrt13-3)/(sqrt13-3)#

#=>tan^2(B/4)=(sqrt13-3)^2/((sqrt13)^2-(3)^2)#

#=>tan^2(B/4)=(sqrt13-3)^2/4#

#=>(tan(B/4))^2=((sqrt13-3)/2)^2...tomangleB <90^circ#

#=>tan(B/4)=(sqrt13-3)/2#

2

#tanx/(1-cotx) + cotx/(1-tanx) = -1#

#=>-tanx/(1-1/tanx) - (1/tanx)/(1-tanx) = 1#

#=>-tan^2x/(tanx-1) - 1/(tanx(1-tanx) )= 1#

#=>tan^2x/(1-tanx) - 1/(tanx(1-tanx) )= 1#

#=>(tan^3x - 1)/(tanx(1-tanx) )= 1#

#=>((tanx - 1)(tan^2x+tanx+1))/(tanx(1-tanx) )= 1#

#tanx=1# will make LHS undefined so neglected.
Hence

#=> - (tan^2x+tanx+1)/tanx= 1#

#=> tan^2x+2tanx+1= 0#

#=>( tanx+1)^2= 0#

#=>tanx=-1#