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Introduction to Trigonometry Test - 31

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Introduction to Trigonometry Test - 31
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  • Question 1
    1 / -0
    The value of sin2600+tan450cosec2450\sin^2 60^0 + \tan 45^0 - \text{cosec}^2 45^0 is
    Solution
    sin2600+tan450cosec2450\sin^2 60^0 + \tan 45^0 - \text{cosec}^2 45^0
    =(32)2+1(2)2=341=14=\left(\dfrac{\sqrt 3}{2}\right)^2+1-(\sqrt 2)^2=\dfrac34-1=\dfrac{-1}{4}
  • Question 2
    1 / -0
    The value of 4(sin4300+cos4600)3(cos2450sin2900) 4(\sin^4 30^0 + \cos^4 60^0)-3 (\cos^2 45^0 \sin^2 90^0) is
    Solution
    4(sin4300+cos4600)3sin2900cos2(450)4(\sin^430^0+\cos^460^0)-3\sin^290^0\cos^2(45^0)

    =4(116+116)3(1)(12)=4(\dfrac{1}{16}+\dfrac{1}{16})-3(1)(\dfrac{1}{2})

    =81632=\dfrac{8}{16}-\dfrac{3}{2}

    =1232=\dfrac{1}{2}-\dfrac{3}{2}

    =1=-1
    Hence answer is 1-1
  • Question 3
    1 / -0
    If acosθb sinθ=ca \cos\theta-b \sin\theta=c, then a sinθ+bcosθa \sin\theta+b\cos\theta equals-
    Solution
    acosθbsinθ=c a\cos\theta - b\sin\theta = c

    Let,asinθ+bcosθ=xa\sin\theta + b\cos\theta = x

    Squaring and adding we get
    a2cos2θ+b2sin2θ2absinθcosθ+a2sin2θ+b2cos2θ+2absinθcosθ=c2+x2a^2\cos^2 \theta+b^2 \sin^2 \theta-2ab\sin \theta \cos \theta+a^2 \sin^2 \theta+b^2 \cos^2 \theta+2ab\sin \theta \cos \theta=c^2+x^2

    a2+b2c2=x2\Rightarrow a^2 + b^2 -c^2 = x^2

    x=±a2+b2c2x=\pm \sqrt{a^2+b^2-c^2}

    Option D is the correct option
  • Question 4
    1 / -0
    The value of sec\sec (900θ)sinθ(90^0 - \theta) \sin \theta is


    Solution
    sec(900θ)sinθ\sec (90^0-\theta)\sin\theta
    =cscθsinθ=\csc\theta\:\sin\theta
    =1sinθsinθ=\dfrac{1}{\sin\theta}\sin\theta
    =1=1
    Hence answer is D
  • Question 5
    1 / -0
    The value of sin2600.cos2300+tan450.cos600sin300\sin^2 60^0.\cos^2 30^0 + \tan 45^0. \cos 60^0 \sin 30^0 is
    Solution

    sin2600cos2300+tan450cos600sin300\sin ^260^0\cos ^230^0+\tan45^0\cos 60^0\sin 30^0

    =sin2600sin2600+(1)cos2600=\sin ^260^0\sin ^260^0+(1)\cos ^260^0

    =sin4600+cos2600=\sin ^460^0+\cos ^260^0

    Substituting the values of sin600\sin 60^0 and cos600\cos 60^0 we get

    $$(\dfrac{\sqrt{3}}{2})\:^4+(\dfrac{1}{2})\:^2$$

    =916+14=\dfrac{9}{16}+\dfrac{1}{4}

    =1316=\dfrac{13}{16}

    Hence answer is A

  • Question 6
    1 / -0
    The value of sin2q.cos2q(sec2q+cosec2q)\sin^2q. \cos^2q (\sec^2q+\text cosec^2q) is
    Solution
    Given: sin2qcos2q(sec2q+cosec2q)\sin^2q \cos^2q (\sec^2q+\text{cosec}^2q)
    =sin2qcos2q(1cos2q+1sin2q)=\sin^2q \cos^2q (\dfrac {1}{\cos^2q}+\cfrac {1}{\sin^2q})

    =sin2qcos2q(sin2q+cos2qsin2qcos2q)=\sin^2q \cos^2q(\dfrac{\sin^2q+\cos^2q}{\sin^2q\cos^2q})

    =sin2q+cos2q=1=\sin^2q+\cos^2q=1
  • Question 7
    1 / -0
    The value of 4cos2600+4tan2450csc23004 \cos^2 60^0 + 4 \tan^2 45^0 - \csc^2 30^0 is
    Solution
    4cos2600+4tan2450csc23004\cos^260^0+4\tan^245^0-\csc^230^0
    $$=4(\dfrac{1}{2})\:^2+4(1)-2^2$$
    =1+44=1+4-4
    =1=1
    Hence answer is C
  • Question 8
    1 / -0
    In the given figure, the side PQPQ (in cm) is

    Solution
    In the given triangle
    AB=BPAB=BP
    Hence, ABP\triangle ABP is an isosceles triangle
    Therefore, PAB=APB\angle PAB=\angle APB
    Hence, BPQ=900(PAB+APB)\angle BPQ=90^0-(\angle PAB+\angle APB)
    =(900150150)=600=(90^0-15^0-15^0)=60^0
    Consider BPQ\triangle BPQ
    cos600=12=PQBP\cos60^0=\dfrac{1}{2}=\dfrac{PQ}{BP}
    =PQ100=\dfrac{PQ}{100}
    Hence, PQ=50PQ=50 cm
    The answer is option B.
  • Question 9
    1 / -0
    The value of sin6q+cos6q+3sin2qcos2q\sin^6q+\cos^6q+3 \sin^2q \cos^2q is-
    Solution
    sin6q+cos6q+3sin2qcos2q\sin^6q+\cos^6q+3 \sin^2q \cos^2q
    (sin2q+cos2q)(sin4q+cos4qsin2qcos2q)+3sin2qcos2q\Rightarrow (\sin^2q+\cos^2q)(\sin^4q+\cos^4q-\sin^2q \cos^2q)+3 \sin^2q \cos^2q
    sin4q+cos4q+2sin2qcos2q\Rightarrow \sin^4q+\cos^4q+2\sin^2q \cos^2q
    (sin2q+cos2q)2=1\Rightarrow (\sin^2q+\cos^2q)^2=1.
  • Question 10
    1 / -0
    The value of sin\sin 450cos450(tan450+cot450)45^0 \cos 45^0 (\tan 45^0 + \cot 45^0) is
    Solution
    sin450cos450(tan450+ cot450)\sin 45^0 \cos 45^0(\tan 45^0+\ cot 45^0)
    =(12)(12)(1+1)=(\dfrac{1}{\sqrt{2}})(\dfrac{1}{\sqrt{2}})(1+1)
    =12(2)=\dfrac{1}{2}(2)
    =1=1
    Hence answer is B
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