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101.

The value of `2(cos^(2)20^(@)+cos^(2)140^(@)+cos^(2)100^(@))` isA. `3//2`B. 3C. 4D. none of these

Answer» Correct Answer - B
`2(cos^(2)20^(@)+cos^(2)140^(@)+cos^(2)100^(@))`
`=3+cos 40^(@)+cos 280^(@)+cos 200^(@)`
`=3+2cos 120^(@) cos 160^(@) + cos 200^(@)`
`= 3-cos 160^(@)+cos 200^(@)`
`=3-cos(360^(@)-200^(@))+cos 200^(@)`
`=3-cos 200^(@)+cos 200^(@)`
= 3
Discussion
102.

If `sin 2theta+sin 2phi=1//2` and `cos2theta+cos 2phi=3//2`, then `cos^(2)(theta-phi)=`A. `3//8`B. `5//8`C. `3//4`D. `5//4`

Answer» Correct Answer - B
Given, `sin 2theta + sin 2phi = 1//2` …(1)
and `cos 2theta + cos 2phi = 3//2` …(2)
Square and adding,
`therefore (sin^(2)2theta + cos^(2) 2theta)+(sin^(2)2phi + cos^(2)2phi)+2[sin 2theta sin 2phi + cos 2theta cos 2phi]=1//4 + 9//4`
`rArr cos 2theta cos 2phi + sin 2theta sin 2phi = 1//4`
`rArr cos (2theta - 2phi)=1//4 rArr cos^(2) (theta - phi) = 5//8`
Discussion
103.

Let `alpha, beta, gamma` be the measures of angle such that `sin alpha+sin beta+sin gamma ge 2`. Then the possible value of `cos alpha + cos beta+cos gamma` isA. 3B. `2.5`C. `2.4`D. 2

Answer» Correct Answer - D
`sin alpha + sin beta + sin gamma ge 2`
or `Sigma sin alpha ge 2`
Squaring both sides, we get
`Sigma sin^(2)alpha + 2 Sigma sin alpha sin beta ge 4`
`rArr 3-Sigma cos^(2)alpha + 2Sigma sin alpha sin beta ge 4`
`rArr 2 Sigma sin alpha sin beta ge 1 1+Sigma cos^(2) alpha`
`rArr 2 Sigma (sin alpha sin beta + cos alpha cos beta) ge 1+(Sigma cos alpha)^(2)` (adding `2Sigma cos alpha cos beta` on both sides)
`rArr (cos alpha + cos beta + cos gamma)^(2)+1le 2[cos (alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)]le 6`
`rArr (cos alpha + cos beta + cos gamma)^(2) le 5`
`rArr (cos alpha + cos beta + cos gamma le sqrt(5)`
Discussion
104.

If `x= cot A + cos A and y= cot A - cos A,` prove that `((x-y)/(x+y))^(2)+((x-y)/(2))^(2)=1.`

Answer» Adding , we get cosec `A=((x+y))/(2)implies sin A=(2) /((x+y)).`
Subtracting , we get , cos `A=(x-y)/(2)`
`therefore sin ^(2)+cos^(2)A=1implies ((2)/(x+y))^(2)+((x-y)/(2))^(2)-1=0.`
Discussion
105.

Value of `(3+cot80^0cot20^0)/(cot80^0+cot20^0)`is equal toA. `cot20^(@)`B. `tan 50^(@)`C. `cot50^(@)`D. `cotsqrt(20^(@))`

Answer» Correct Answer - B
`([3+(cos 80^(@)cos20^(@))/(sin80^(@)sin20^(@))])/([(cos 80^(@))/(sin80^(@))+(cos20^(@))/(sin20^(@))])`
`=([2sin80^(@)sin20^(@)+(cos 80^(@)cos20^(@)sin80^(@)sin20^(@))])/(sin20^(@)cos80^(@)+cos20^(@)sin80^(@))`
Discussion
106.

If `tanalpha`is equal to the integral solution of the inequality `4x^2-16 x+15

Answer» Correct Answer - D
We have `4x^(2)-16x+15lt0`
`rArr (3)/(2)ltxlt(5)/(2)`
Therefore, the integral solution of
`4x^(2)-16x+15lt0` is `x=2`.
Thus, `tan alpha=2`. It is given that `cos beta=tan45^(@)=1`.
`therefore sin(alpha+beta)sin(alpha-beta)=sin^(2)alpha-sin^(2)beta`.
`(1)/(1+cot^(2)alpha)(1-cos^(2)beta)`
`=(1)/(1+(1)/(4))-0=(4)/(5)`
Discussion
107.

In the figure of `Delta PQR , angle P=theta^(@) and angle R= phi ^(@)`. find `(1) (sqrt(x+1))cot phi` `(ii) (sqrt(x^(3)+x^(2))) tan theta` `(iii) cos theta `

Answer» Correct Answer - `(i) (x)/(2)" " (ii) (x^(2))/(2)" " (iii) (2sqrt(x+1))/((x+2))`
`(i) cot phi =(QR)/(PQ)=(x)/(2sqrt(x+1))implies (sqrt(x)+1)cot phi =(x)/(2).`
`(ii) tan theta =(QR)/(PQ)=(x)/(2sqrt(x+1))=(x^(2))/(2sqrt(x^(3)+x^(2)))implies (sqrt(x^(3)+x^(2))) tan theta =(x^(2))/(2).`
`(iii) cos theta =(PQ)/(PR)=(2sqrt(x+1))/((x+2)).`
Discussion
108.

If `alpha` and `beta` are acute such that `alpha+beta` and `alpha-beta` satisfy the equation `tan^(2)theta-4tan theta+1=0`, then `(alpha, beta`) =A. `(30^(@), 60^(@))`B. `(45^(@),45^(@))`C. `(45^(@),30^(@))`D. `(60^(@),45^(@))`

Answer» Correct Answer - C
`tan^(2)theta-4 tan theta + 1=0`
`rArr tan theta = (4 pm sqrt(12))/(2)=2pm sqrt(3)`
`rArr tan (alpha+beta)=2+sqrt(3)` and `tan (alpha-beta)=2-sqrt(3)`
`rArr alpha+beta=75^(@)` and `alpha-beta =15^(@)`
`rArr alpha = 45^(@)` and `beta = 30^(@)`
Discussion
109.

Let `f(n)=2cosn xAAn in N ,`then `f(1)f(n+1)-f(n)`is equal to`f(n+3)`(b) `f(n+2)``f(n+1)f(2)`(d) `f(n+2)f(2)`A. `f(n+3)`B. `f(n+2)`C. `f(n+1)f(2)`D. `f(n+2)f(2)`

Answer» Correct Answer - B
`f(n)=2cos nx`
`rArr f(1)f(n+1)-f(n)`
`=4cos x cos(n+1)x-2cos nx`
`=2[2cos (n+1)x cos x-cos nx]`
`=2[cos(n+2)x+cos nx-cosnx]`
`=2cos(n+2)x=f(n+2)`.
Discussion
110.

Let `x, y, in R` satisfy the condition such that sin x sin y + 3 cos y +4 sin y cos `x=sqrt(26)`. The value of `tan^(2)x + cot^(2)y` is equal toA. `9xx17`B. 205C. `(1)/(16)+(9)/(17)`D. none of these

Answer» Correct Answer - C
`sin x sin y + 3 cos y + 4 sin y cos x = sqrt(26)`
`therefore 3 cos y + (sin x + 4 cos x) sin y = sqrt(26)`
`therefore 3 cos y + (sin x + 4 cos x)sin y`
`le sqrt(9+(sin x+4 cos x)^(2))`
`le sqrt(9+1+16)`
`= sqrt(26)`
`therefore sin x sin y + 3 cos y + 4 sin y cos x = sqrt(26)`
`rArr sin x sin y = (cos y)/(3) = (sin y cos x)/(4)`
`rArr 3 tna y = cosec x` and `tan x = 1//4`
`rArr 9 tan^(2)y = cosec^(2)x=(1+cot^(2)x)=17`
`rArr tan^(2) x + cot^(2)y=(1)/(16)+(9)/(17)`
Discussion
111.

If `sintheta_1sintheta_2-costheta_1costheta_2+1=0,`then the value of `tan((theta_1)/2)cot((theta_2)/2)`is equal to`-1`(b) 1(c) 2 (d) `-2`A. `a^(2)+b^(2)ge4`B. `a^(2)+b^(2)le4`C. `a^(2)+b^(2)ge3`D. `a^(2)+b^(2)le2`

Answer» Correct Answer - B
`sintheta_(1)-sin theta_(2)=a,cos theta_(1)+cosheta_(2)=b`
`rArr a^(2)+b^(2)=2+2cos(theta_(1)+theta_(2))`
`rArr 0lea^(2)+b^(2)le4`
Discussion
112.

If `costheta_1 2costheta_2,`then `tan(theta_(1-theta_2))/2tan(theta_1+theta_2)/2`is equal to`1/3`(b) `-1/3``1`(d) `-1`A. `(1)/(3)`B. `-(1)/(3)`C. 1D. `-1`

Answer» Correct Answer - B
`sin27^(@)-sin63^(@)=-2cos45^(@)sin18^(@)`
`rArr tan(x+y)=((4sqrt(5))/(3sqrt(3)))/(1-(15)/(27))=(4sqrt(5)xx27)/(12xx3sqrt(3))=sqrt(13)`
Discussion
113.

If `alpha,beta,gamma`are ccute angles and `costheta=sinbeta//sinalpha,cosvarphi=singammasinalphaa n dcos(theta-varphi)=sinbetasingamma`, then the value of `tan^2alpha-tan^2beta-tan^2gamma`is equal to`-1`(b) `0`(c) `1`(d) 2A. `-1`B. 0C. 1D. 2

Answer» Correct Answer - B
From the third relation, we get
`cos theta cos phi + sin theta sin phi = sin beta sin gamma`
`rArr sin^(2)theta sin^(2)phi=(cos theta cos phi - sin beta sin gamma)^(2)`
`rArr (1-(sin^(2)beta)/(sin^(2)alpha))(1-(sin^(2)gamma)/(sin^(2)alpha))=((sin beta sin gamma)/(sin^(2)alpha)-sin beta sin gamma)^(2)`
`rArr (sin^(2)alpha-sin^(2)beta)(sin^(2)alpha-sin^(2)gamma)=sin^(2)beta sin^(2)gamma(1-sin^(2)alpha)^(2)`
`rArr sin^(4)alpha(1-sin^(2)beta sin^(2)sin^(2)gamma)-sin^(2)alpha(sin^(2)beta + sin^(2)gamma-2 sin^(2)beta sin^(2)gamma)=0`
`therefore sin^(2)alpha=(sin^(2)beta + sin^(2)gamma -2sin^(2)beta sin^(2)gamma)/(1-sin^(2)beta si8n^(2)gamma)`
and `cos^(2)alpha=(1-sin^(2)beta-sin^(2)gamma + sin^(2)beta sin^(2)gamma)/(1-sin^(2)beta sin^(2)gamma)`
`rArr tan^(2)alpha=(sin^(2)beta -sin^(2)beta sin^(2)gamma+sin^(2)gamma-sin^(2)beta sin^(2)gamma)/(cos^(2)beta - sin^(2)gamma(1-sin^(2)beta))`
`=(sin^(2)beta cos^(2)gamma + cos^(2)beta sin^(2)gamma)/(cos^(2)beta cos^(2)gamma)`
`= tan^(2) bet + tan^(2) gamma`
`rArr tan^(2) alpha - tan^(2) beta - tan^(2) gamma = 0`
Discussion