01Medium45×since 2004Q5635If cos(α\alphaα + β\betaβ) = 3/5 ,sin ( α\alphaα - β\betaβ) = 5/13 and 0 < α,β\alpha , \betaα,β < π4\pi \over 44π, then tan(2α\alphaα) is equal to :A21/16B63/52C33/52D63/16Check answerSkip
02Medium45×since 2004Q5636If 0 < x, y < π\piπ and cosx + cosy −-− cos(x + y) = 32{3 \over 2}23, then sinx + cosy is equal to :A1+32{{1 + \sqrt 3 } \over 2}21+3B12{{1 \over 2}}21C32{{\sqrt 3 } \over 2}23D1−32{{1 - \sqrt 3 } \over 2}21−3Check answerSkip
03Medium45×since 2004Q5637If tan(π9),x,tan(7π18)\tan \left( {{\pi \over 9}} \right),x,\tan \left( {{{7\pi } \over {18}}} \right)tan(9π),x,tan(187π) are in arithmetic progression and tan(π9),y,tan(5π18)\tan \left( {{\pi \over 9}} \right),y,\tan \left( {{{5\pi } \over {18}}} \right)tan(9π),y,tan(185π) are also in arithmetic progression, then ∣x−2y∣|x - 2y|∣x−2y∣ is equal to :A4B3C0D1Check answerSkip
04Medium45×since 2004Q5638If cotα\alphaα = 1 and secβ\betaβ = −53- {5 \over 3}−35, where π<α<3π2\pi < \alpha < {{3\pi } \over 2}π<α<23π and π2<β<π{\pi \over 2} < \beta < \pi2π<β<π, then the value of tan(α+β)\tan (\alpha + \beta )tan(α+β) and the quadrant in which α\alphaα + β\betaβ lies, respectively are :A−17- {1 \over 7}−71 and IV^th quadrantB7 and I^st quadrantC−-−7 and IV^th quadrantD17{1 \over 7}71 and I^st quadrantCheck answerSkip
05Hard45×since 2004Q5639If tanA=1x(x2+x+1),tanB=xx2+x+1\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}tanA=x(x2+x+1)1,tanB=x2+x+1x and tanC=(x−3+x−2+x−1)1/2,0<A,B,C<π2\tan \mathrm{C}=\left(x^{-3}+x^{-2}+x^{-1}\right)^{1 / 2}, 0<\mathrm{A}, \mathrm{B}, \mathrm{C}<\frac{\pi}{2}tanC=(x−3+x−2+x−1)1/2,0<A,B,C<2π, then A+B\mathrm{A}+\mathrm{B}A+B is equal to :AC\mathrm{C}CBπ−C\pi-Cπ−CC2π−C2 \pi-C2π−CDπ2−C\frac{\pi}{2}-\mathrm{C}2π−CCheck answerSkip
06Medium45×since 2004Q5640For α,β∈(0,π/2)\alpha, \beta \in(0, \pi / 2)α,β∈(0,π/2), let 3sin(α+β)=2sin(α−β)3 \sin (\alpha+\beta)=2 \sin (\alpha-\beta)3sin(α+β)=2sin(α−β) and a real number kkk be such that tanα=ktanβ\tan \alpha=k \tan \betatanα=ktanβ. Then, the value of kkk is equal toA5B−-−2/3C−-−5D2/3Check answerSkip
07Medium45×since 2004Q5641The number of solutions, of the equation esinx−2e−sinx=2e^{\sin x}-2 e^{-\sin x}=2esinx−2e−sinx=2, is :A0B1C2Dmore than 2Check answerSkip
08Medium45×since 2004Q5642If 0<x<π0 < x < \pi0<x<π and cosx+sinx=12,\cos x + \sin x = {1 \over 2},cosx+sinx=21, then tanx\tan xtanx is :A(1−7)4{{\left( {1 - \sqrt 7 } \right)} \over 4}4(1−7)B(4−7)3{{\left( {4 - \sqrt 7 } \right)} \over 3}3(4−7)C−(4+7)3- {{\left( {4 + \sqrt 7 } \right)} \over 3}−3(4+7)D(1+7)4{{\left( {1 + \sqrt 7 } \right)} \over 4}4(1+7)Check answerSkip
09Medium45×since 2004Q5644Let fk(x)=1k(sinkx+coskx)f_k\left( x \right) = {1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)fk(x)=k1(sinkx+coskx) where x∈Rx \in Rx∈R and k≥ 1.k \ge \,1.k≥1. Then f4(x)−f6(x) {f_4}\left( x \right) - {f_6}\left( x \right)\,\,f4(x)−f6(x) equals :A14{1 \over 4}41B112{1 \over 12}121C16{1 \over 6}61D13{1 \over 3}31Check answerSkip
10Hard45×since 2004Q5645For any θ∈(π4,π2)\theta \in \left( {{\pi \over 4},{\pi \over 2}} \right)θ∈(4π,2π), the expression 3(cosθ−sinθ)43{(\cos \theta - \sin \theta )^4}3(cosθ−sinθ)4+6(sinθ+cosθ)2+4sin6θ+ 6{(\sin \theta + \cos \theta )^2} + 4{\sin ^6}\theta+6(sinθ+cosθ)2+4sin6θ equals :A13 – 4 cos²θ\thetaθ + 6sin²θ\thetaθcos²θ\thetaθB13 – 4 cos⁶θ\thetaθC13 – 4 cos²θ\thetaθ + 6cos²θ\thetaθD13 – 4 cos⁴θ\thetaθ + 2sin²θ\thetaθcos²θ\thetaθCheck answerSkip