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JEE • MATHEMATICS • TRIGONOMETRY RATIOS

Trigonometry ratios practice · Mixed difficulty

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211 Medium
211 of 394
Application +4 | -1
If tan⁡α=(1+2−x)−1,tan⁡β=(1+2x+1)−1\tan \alpha=\left(1+2^{-x}\right)^{-1}, \tan \beta=\left(1+2^{x+1}\right)^{-1}, then α+β\alpha+\beta equals
View answer and explanation
Correct option: B
π/4\pi / 4
Option B (π4\frac{{\pi}}{{4}}) is correct because by substituting t=2xt = 2^x and using the tangent addition formula, tan⁡(α+β)\tan(\alpha + \beta) simplifies to 1, which means α+β=π4\alpha + \beta = \frac{{\pi}}{{4}}.
212 Medium
212 of 394
Analyzing +4 | -1
If in a △ABC,2cos⁡Aa+cos⁡Bb+2cos⁡Cc=abc+bca\triangle A B C, \frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}, then ∠A\angle A is equal to
View answer and explanation
Correct option: A
90∘90^{\circ}
Option A (90∘90^{\circ}) is correct because simplifying the given equation using the sine rule and cosine rule leads to the conclusion that angle A is 90∘90^{\circ}.
213 Easy
213 of 394
Analyzing +4 | -1
For θ>π/3,\mathrm{\theta } > \mathrm{\pi }/3, the value of f(θ)=sec⁡2θ+cos⁡2θf\left( \mathrm{\theta } \right) = {\sec ^2}\mathrm{\theta } + {\cos ^2}\mathrm{\theta } always lies in the interval
View answer and explanation
Correct option: D
[2,∞)\left[ {2,\infty } \right)
Option D ([2,∞)[2,\infty)) is correct because the minimum value of the function is 2, and it can go to infinity.
214 Easy
214 of 394
Application +4 | -1
For x∈IR,  3cos⁡  (4x−5)+4x \in IR,\,\,3\cos \,\,\left( {4x - 5} \right) + 4 lies in the interval
View answer and explanation
Correct option: A
[1,7]\left[ {1,\mathrm{ }7} \right]
Option A ([1, 7]) is correct because the minimum value of 3cos⁡(4x−5)3\cos(4x-5) is -3 and the maximum value is 3, so the range of 3cos⁡(4x−5)+43\cos(4x-5) + 4 is [1, 7].
215 Medium
215 of 394
Application +4 | -1
If cos⁡A3=cos⁡B4=15,−π2<(A,B)<0\frac{\cos A}{3}=\frac{\cos B}{4}=\frac{1}{5},-\frac{\pi}{2} < (A, B) < 0, then value of 2sin⁡A+4sin⁡B2 \sin A+4 \sin B is
View answer and explanation
Correct option: C
-4
Option C (-4) is correct because given the conditions, we find sin⁡A=−45\sin A = -\frac{4}{5} and sin⁡B=−35\sin B = -\frac{3}{5}. Substituting these values into 2sin⁡A+4sin⁡B2 \sin A + 4 \sin B gives 2(−45)+4(−35)=−85−125=−42(-\frac{4}{5}) + 4(-\frac{3}{5}) = -\frac{8}{5} - \frac{12}{5} = -4.
216 Medium
216 of 394
Understanding +4 | -1
The value of cos⁡10∘−sin⁡10∘\cos 10^\circ - \sin 10^\circ is
View answer and explanation
Correct option: A
Positive
Option A (Positive) is correct because for angles between 0 and 45 degrees, the cosine function is greater than the sine function.
217 Easy
217 of 394
Application +4 | -1
If cos⁡  θ=817\cos \,\,\theta = \frac{8}{{17}} and θ\theta lies in the I st quadrant, then the value of cos⁡  (30∘+θ)+cos⁡  (45∘−θ)+cos⁡  (120∘−θ)\cos \,\,\left( {30^\circ + \theta } \right) + \cos \,\,\left( {45^\circ - \theta } \right) + \cos \,\,\left( {120^\circ - \theta } \right) is
View answer and explanation
Correct option: A
2317  (3−12+12)\frac{{23}}{{17}}\,\,\left( {\frac{{\sqrt 3 - 1}}{2} + \frac{1}{{\sqrt 2 }}} \right)
Option A is correct because it represents the simplified expression after applying trigonometric sum-to-product formulas and substituting the given values.
218 Medium
218 of 394
Application +4 | -1
The value of cos⁡(π/13)+cos⁡(2π/13)+….+cos⁡(12π/13)\cos (\pi / 13)+\cos (2 \pi / 13)+\ldots .+\cos (12 \pi / 13) is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the terms in the cosine series cancel each other out due to the symmetry around π\pi.
219 Easy
219 of 394
Application +4 | -1
cos⁡(540∘−θ)−sin⁡(630∘−θ)\cos \left(540^{\circ}-\theta\right)-\sin \left(630^{\circ}-\theta\right) is equal to
View answer and explanation
Correct option: A
0
The correct answer is A (0) because cos⁡(540∘−θ)−sin⁡(630∘−θ)\cos(540^{\circ} - \theta) - \sin(630^{\circ} - \theta) simplifies to −cos⁡θ−(−cos⁡θ)=0-\cos \theta - (-\cos \theta) = 0.
220 Easy
220 of 394
Application +4 | -1
The value of 6(sin⁡6θ+cos⁡6θ)−9  (sin⁡4θ+cos⁡4θ)+4  6({\sin ^6}\mathrm{\theta } + {\cos ^6}\mathrm{\theta }) - 9\mathrm{\;}({\sin ^4}\mathrm{\theta } + {\cos ^4}\mathrm{\theta }) + 4\mathrm{\;} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the expression simplifies to 1 using trigonometric identities.
221 Medium
221 of 394
Analyzing +4 | -1
If a and b are positive numbers such that a>b,a > b, then the minimum value of a  sec  θ−btan⁡θ(0<θ<π2)\mathrm{a\;sec}\;\theta - b\tan \theta \left( {0 < \theta < \frac{\pi }{2}} \right) is
View answer and explanation
Correct option: D
a2−b2\sqrt {{a^2} - {b^2}}
Option D (a2−b2\sqrt{{{{a^2} - {b^2}}}}) is correct because it represents the minimum value of the expression, found using differentiation and trigonometric identities.
222 Medium
222 of 394
Understanding +4 | -1
Which of the following relations is possible?
View answer and explanation
Correct option: B
  tan  θ=1002\mathrm{\;tan}\,\,\mathrm{\theta } = {100^2}
Option B (  tan  θ=1002\;tan\,\,\theta = {100^2}) is correct because the range of the tangent function is all real numbers.
223 Easy
223 of 394
Analyzing +4 | -1
If (sec⁡A−tan⁡A)  (sec⁡B−tan⁡   B)  (sec⁡C−tan⁡C)=(sec⁡A+tan⁡A)  (sec⁡B+tan⁡B)  (sec⁡C+tan⁡C)\left( {\sec A - \tan A} \right)\,\,\left( {\sec B - \tan \;\,B} \right)\,\,\left( {\sec C - \tan C} \right) = \left( {\sec A + \tan A} \right)\,\,\left( {\sec B + \tan B} \right)\,\,\left( {\sec C + \tan C} \right) then each side is equal to
View answer and explanation
Correct option: D
±1\pm 1
Option D (±1\pm 1) is correct because the given equation implies that each side is equal to either 1 or -1.
224 Easy
224 of 394
Application +4 | -1
Let fk(x)=1k(sin⁡kx+cos⁡kx){f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right) for k=1,2,3...k = 1,2,3... Then, for all x∈Rx \in R , the value of f4(x)−f6(x){f_4}\left( x \right) - {f_6}\left( x \right) is equal to
View answer and explanation
Correct option: A
112\frac{1}{{12}}
Option A (112\frac{1}{{12}}) is correct because it's the result of simplifying the expression f4(x)−f6(x){f_4}(x) - {f_6}(x) using trigonometric identities.
225 Easy
225 of 394
Understanding +4 | -1
If sin⁡  θ1+sin⁡  θ2+sin⁡  θ3=3\sin \,\,{\theta _1} + \sin \,\,{\theta _2} + \sin \,\,{\theta _3} = 3, then cos⁡  θ1+cos⁡  θ2+cos⁡  θ3=\cos \,\,{\theta _1} + \cos \,\,{\theta _2} + \cos \,\,{\theta _3} =
View answer and explanation
Correct option: D
0\mathrm{0}
Option D (0) is correct because the maximum value of sin⁡θ\sin \theta is 1, which implies θ1=θ2=θ3=π2\theta_1 = \theta_2 = \theta_3 = \frac{\pi}{2}, making the sum of cosines zero.

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226 Easy
226 of 394
Analyzing +4 | -1
If a  sin⁡2  x+b  cos⁡2  x=c,    b  sin⁡2  y+a  cos⁡2  y=da\,\,{\sin ^2}\,\,x + b\,\,{\cos ^2}\,\,x = c,\,\,\;b\,\,{\sin ^2}\,\,y + a\,\,{\cos ^2}\,\,y = d and a  tan⁡  x=b  tan⁡  ya\,\,\tan \,\,x = b\,\,\tan \,\,y, then a2b2\frac{{{a^2}}}{{{b^2}}} is equal to

View answer and explanation
Correct option: B
(a−d)  (c−a)(b−c)  (d−b)\frac{{\left( {a - d} \right)\,\,\left( {c - a} \right)}}{{\left( {b - c} \right)\,\,\left( {d - b} \right)}}
Option B ((a−d)(c−a)(b−c)(d−b)\frac{{\left( {a - d} \right)\left( {c - a} \right)}}{{\left( {b - c} \right)\left( {d - b} \right)}}) is correct because it is derived from manipulating the given equations and using trigonometric identities.
227 Medium
227 of 394
Analyzing +4 | -1
3 (sin⁡  x−cos⁡  x)4+6  (sin⁡  x+cos⁡  x)2+4    (sin⁡6  x+cos⁡6  x)3\,{(\sin \,\,x - \cos \,\,x)^4} + 6\,\,{\left( {\sin \,\,x + \cos \,\,x} \right)^2} + 4\,\,\;\left( {{{\sin }^6}\,\,x + {{\cos }^6}\,\,x} \right) is equal to
View answer and explanation
Correct option: C
13\mathrm{13}
Option C (13) is correct because after simplification using trigonometric identities, the expression evaluates to 13.
228 Easy
228 of 394
Application +4 | -1
If cos⁡θ=cos⁡αcos⁡β\cos \theta=\cos \alpha \cos \beta, then tan⁡θ+α2tan⁡θ−α2\tan \frac{\theta+\alpha}{2} \tan \frac{\theta-\alpha}{2} is equal to
View answer and explanation
Correct option: D
cot⁡2(β/2)\cot ^2(\beta / 2)
Option B (tan⁡2(β/2)\tan ^2(\beta / 2)) is correct because the given trigonometric expression simplifies to tan⁡2(β/2)\tan^2(\beta/2) using trigonometric identities.
229 Easy
229 of 394
Application +4 | -1
If Sn=cos⁡n  θ+sin⁡n  θ  {S_n} = {\cos ^n}\,\,\theta + {\sin ^n}\,\,\theta \;, then the value of 3S4−2S63{S_4} - 2{S_6} is given by
View answer and explanation
Correct option: C
1
Option C (1) is correct because substituting the given expression and simplifying using trigonometric identities results in 1.
230 Easy
230 of 394
Application +4 | -1
The function f(x)=acos⁡x+btan⁡x+xf\left( x \right) = a\cos x + b\tan x + x has extreme values at x=0x = 0 and x=π6x = \frac{\pi }{6}, then
View answer and explanation
Correct option: A
a=−23,b=−1a = - \frac{2}{3},b = - 1
Option C (0,0) is correct because when the tangent is perpendicular to the x-axis, the slope is infinite, implying t=0, which gives the point (0,0).
231 Easy
231 of 394
Application +4 | -1
If cos⁡θ+sin⁡ϕ=m\cos \theta+\sin \phi=m and sin⁡θ+cos⁡ϕ=n\sin \theta+\cos \phi=n, then 12(m2+n2−2)\frac{1}{2}\left(m^2+n^2-2\right) is equal to
View answer and explanation
Correct option: B
sin⁡(θ+ϕ)\sin (\theta+\phi)
Option B (sin⁡(θ+ϕ)\sin (\theta+\phi)) is correct because simplifying the expression 12(m2+n2−2)\frac{{1}}{{2}}(m^2 + n^2 - 2) using the given equations results in sin⁡(θ+ϕ)\sin (\theta + \phi).
232 Medium
232 of 394
Application +4 | -1
tan⁡100∘+tan⁡125∘+tan⁡100∘tan⁡125∘=\tan 100^{\circ}+\tan 125^{\circ}+\tan 100^{\circ} \tan 125^{\circ}=
View answer and explanation
Correct option: D
1
Option D (1) is correct because using the tangent addition formula, the expression simplifies to 1.
233 Medium
233 of 394
Analyzing +4 | -1
If mm and MM are the minimum and the maximum values of 4+12sin⁡22x−2cos⁡4x,x∈R4+\frac{1}{2} \sin ^{2} 2 x-2 \cos ^{4} x, x \in R, then M−mM-m is equal to:
View answer and explanation
Correct option: B
154\frac{15}{4}
Option B (15/4) is correct because by completing the square and finding the maximum and minimum values of the given expression, the difference M-m equals 15/4.
234 Easy
234 of 394
Analyzing +4 | -1
If sin⁡4α+4cos⁡4 β+2=42  sin⁡α  cos⁡β ;   α,β  ∈  [0, π],{\sin ^4}\alpha + 4{\cos ^4}\,\beta + 2 = 4\sqrt 2 \,\,\sin \alpha \,\,\cos \beta \,;\,\,\,\alpha ,\beta \,\, \in \,\,\left[ {0,\,\pi } \right], then cos⁡(α+β)−cos⁡(α−β)\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right) is equal to
View answer and explanation
Correct option: C
−2- \sqrt 2
Option C (−2- \sqrt 2) is correct because by applying AM >= GM inequality, we find the values of sin⁡α\sin \alpha and cos⁡β\cos \beta, and then calculate the expression cos⁡(α+β)−cos⁡(α−β)\cos (\alpha + \beta ) - \cos (\alpha - \beta ).
235 Medium
235 of 394
Application +4 | -1
Which of the following is rational?
View answer and explanation
Correct option: C
sin⁡15∘cos⁡15∘\sin 15^{\circ} \cos 15^{\circ}
Option C (sin⁡15∘cos⁡15∘\sin 15^{\circ} \cos 15^{\circ}) is correct because sin⁡15∘cos⁡15∘=1/4\sin 15^{\circ} \cos 15^{\circ} = 1/4, which is a rational number.
236 Easy
236 of 394
Application +4 | -1
If sin⁡x+cos⁡y=a\sin x+\cos y=a and cos⁡x+sin⁡y=b\cos x+\sin y=b, then tan⁡x−y2\tan \frac{x-y}{2} is equal to
View answer and explanation
Correct option: D
a−ba+b\frac{a-b}{a+b}
Option D (a−ba+b\frac{{a-b}}{{a+b}}) is correct because it is derived by using sum-to-product trigonometric identities and then applying the tangent half-angle formula.
237 Medium
237 of 394
Application +4 | -1
If in a △ABC,(a+b+c)(b+c−a)=λbc\triangle A B C,(a+b+c)(b+c-a)=\lambda b c, then
View answer and explanation
Correct option: D
0<λ<40<\lambda<4
Option D (0<λ<40<\lambda<4) is correct because the given equation can be transformed to relate λ\lambda to cos⁡A\cos A, and the range of cos⁡A\cos A restricts the possible values of λ\lambda.
238 Easy
238 of 394
Application +4 | -1
The value of cos⁡2π7+cos⁡4π7+\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+ cos⁡6π7+cos⁡7π7\cos \frac{6 \pi}{7}+\cos \frac{7 \pi}{7} is
View answer and explanation
Correct option: D
−3/2-3 / 2
Option D (-3/2) is correct because by using the sum-to-product trigonometric identities, the expression simplifies to -3/2.
239 Easy
239 of 394
Application +4 | -1
With usual notations in a △ABC\triangle A B C, b2−c2asec⁡A+c2−a2bsec⁡B+a2−b2csec⁡C\frac{b^2-c^2}{a \sec A}+\frac{c^2-a^2}{b \sec B}+\frac{a^2-b^2}{c \sec C} is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because by applying the Law of Cosines and simplifying, the expression evaluates to zero.
240 Medium
240 of 394
Analyzing +4 | -1
Let θ∈(0,   π4)\theta \in \left( {0,\;\,\frac{\pi }{4}} \right) and t1=(tan⁡  θ)tan⁡θ,   t2=(tan⁡  θ)cot⁡θ,   t3=(cot⁡  θ)tan⁡θ{t_1} = {\left( {\tan \,\,\theta } \right)^{\tan \theta }},\,\;{t_2} = {\left( {\tan \,\,\theta } \right)^{\cot \theta }},\;\,{t_3} = {\left( {\cot \,\,\theta } \right)^{\tan \theta }} and t4=(cot⁡  θ)tan⁡θ{t_4} = {\left( {\cot \,\,\theta } \right)^{\tan \theta }} , then
View answer and explanation
Correct option: B
t4>t3>t1>t2{t_4} > {t_3} > {t_1} > {t_2}
Option B (t4>t3>t1>t2{t_4} > {t_3} > {t_1} > {t_2}) is correct because when θ∈(0,π4)\theta \in (0, \frac{\pi}{4}), tan⁡θ<1\tan \theta < 1 and cot⁡θ>1\cot \theta > 1, leading to the given inequality.
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