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Mathematics Test - 16

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Mathematics Test - 16
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  • Question 1
    1 / -0

    If 7sinα = 24 cosα; 0 < α < π/2, then the value of 14 tanα−75cosα − 7secα is equal to:

    Solution
    Given, \(7 \sin \alpha=24 \cos \alpha\)
    \(\Rightarrow \frac{\sin \alpha}{\cos \alpha}=\tan \alpha=\frac{24}{7}=\frac{P}{B}\)
    Hence, \(P=24\) and \(B=7 \mathrm{cm}\)
    \(\therefore H=\sqrt{P^{2}+B^{2}}\)
    \(\Rightarrow H=\sqrt{\left.(24)^{2}+7\right)^{2}}\)
    \(\Rightarrow H=\sqrt{625}\)
    \(\Rightarrow H=25\)
    \(\cos \alpha=\frac{B}{H}=\frac{7}{25}\)
    \(\sec \alpha=\frac{H}{B}=\frac{25}{7}\)
    Hence, \(14 \tan \alpha-75 \cos \alpha-7 \sec \alpha\)
    \(=14 \times \frac{24}{7}-75 \times \frac{7}{25}-7 \times \frac{25}{7}\)
    \(=48-21-25=2\)
    Hence, the correct option is B.
  • Question 2
    1 / -0

    The value of \(\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}\) is :

    Solution
    \(\cos \left(40^{\circ}+\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}\)
    \(=\sin \left[90^{\circ}-\left(40^{\circ}+\theta\right)\right]-\sin \left(50^{\circ}-\theta\right)+\frac{\cos ^{2}\left(90^{\circ}-50^{\circ}\right)+\cos ^{2}\left(90^{\circ}-40^{\circ}\right)}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}\)
    \(=\sin \left(50^{\circ}-\theta\right)-\sin \left(50^{\circ}-\theta\right)+\frac{\sin ^{2} 50^{\circ}+\sin ^{2} 40^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}\)
    \(=0+1=1\)
    Hence, the correct option is A.
  • Question 3
    1 / -0

    If sinA + cosA = m and sin3 A + cos3 A = n, then

    Solution
    \(\sin ^{3} A+\cos ^{3} A=n\)
    \(\sin A+\cos A=m\)
    \((\sin A+\cos A)^{3}=m^{3}\)
    \(\sin ^{3} A+\cos ^{3} A+3 \sin A \cos A(\sin A+\cos A)=m^{3}\)
    \(n+3 m \sin A \cos A=m^{3} \ldots \ldots .(i)\)
    \((\sin A+\cos A)^{2}=m^{2}\)
    \(\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A=m^{2}\)
    \(1+2 \sin A \cos A=m^{2}\)
    \(\sin A \cos A=\frac{m^{2}-1}{2}\)
    Substituting in \((i)\) \(\Rightarrow n+3 m\left(\frac{m^{2}-1}{2}\right)=m^{3}\)
    \(\Rightarrow 2 n+3 m^{3}-3 m=2 m^{3}\)
    \(\Rightarrow m^{3}-3 m+2 n=0\)
    Hence, the correct option is C.
  • Question 4
    1 / -0
    In an isosceles triangle ABC; AB = AC = 10 cm and BC = 18 cm. Find the value of :
    sin2B + cos2C
    Solution
    Given, \(A B=A C=10\) and \(B C=18 \mathrm{cm}\)
    \(\cos B=\frac{A B^{2}+B C^{2}-A C^{2}}{2 A B \times B C}\)
    \(\cos B=\frac{10^{2}+18^{2}-10^{2}}{2 \times 10 \times 18}\)
    \(\cos B=\frac{9}{10}\)
    \(\cos B=\frac{B}{H}=\frac{9}{10}\)
    Using Pythagoras Theorem, \(H^{2}=P^{2}+B^{2}\)
    \(10^{2}=P^{2}+9^{2}\)
    \(P=\sqrt{19}\)
    \(\sin B=\frac{P}{H}=\frac{\sqrt{19}}{10}\)
    \(\mathrm{Now}, \cos C=\frac{A C^{2}+B C^{2}-A B^{2}}{2 A C \times B C}\)
    \(\cos C=\frac{10^{2}+18^{2}-10^{2}}{2 \times 10 \times 18}\)
    \(\cos C=\frac{9}{10}\)
    Thus, \(\sin ^{2} B+\cos ^{2} C=\left(\frac{\sqrt{19}}{10}\right)^{2}+\left(\frac{9}{10}\right)^{2}\)
    \(\sin ^{2} B+\cos ^{2} C=\frac{19}{100}+\frac{81}{100}\)
    \(\sin ^{2} B+\cos ^{2} C=\frac{100}{100}=1\)
    Hence, the correct option is C.
  • Question 5
    1 / -0
    \(a \sin x=b \cos x=\frac{2 c \tan x}{1-\tan ^{2} x}\) and \(\left(a^{2}-b^{2}\right)=k c^{2}\left(a^{2}+b^{2}\right)\)
    then \(k=?\)
    Solution
    Given that:
    \(a \sin x=b \cos x=\frac{2 \operatorname{ctan} x}{1-\tan ^{2} x}\)
    and \(\left(a^{2}-b^{2}\right)^{2}=k c^{2}\left(a^{2}+b^{2}\right)\)
    To find:
    \(k=?\)
    Solution:
    \(a \sin x=b \cos x\)
    or, \(\tan x=\frac{b}{a}\)
    \(a \sin x=\frac{2 \operatorname{ctan} x}{1-\tan ^{2} x}\)
    \(2 c \times \frac{b}{a}\)
    or, \(a \sin x=\frac{a}{1-\left(\frac{b}{a}\right)^{2}}\)
    or, \(a \times \frac{b}{\sqrt{a^{2}+b^{2}}}=\frac{2 a b c}{a^{2}-b^{2}}\)
    or, \(\frac{1}{\sqrt{a^{2}+b^{2}}}=\frac{2 c}{a^{2}-b^{2}}\)
    or, \(\left(a^{2}-b^{2}\right)=2 c \sqrt{a^{2}+b^{2}}\)
    Squaring both sides we get,
    or, \(\left(a^{2}-b^{2}\right)^{2}=4 c^{2}\left(a^{2}+b^{2}\right)\)
    So, value of \(k=4\)
    Hence, the correct option is D.
  • Question 6
    1 / -0
    \(x=\frac{\sin ^{3} p}{\cos ^{2} p}, y=\frac{\cos ^{3} p}{\sin ^{2} p}\) and \(\sin p+\cos p=\frac{1}{2}\) then \(x+y=\)
    Solution
    \(\sin p+\cos p=\frac{1}{2} \Rightarrow(\sin p+\cos p)^{2}=\frac{1}{4}\)
    \(1+2 \sin p \cdot \cos p=\frac{1}{4} \Rightarrow \sin p \cos p=-\frac{3}{8}\)
    Therefore, \(x+y=\frac{\sin ^{3} p}{\cos ^{2} p}+\frac{\cos ^{3} p}{\sin ^{2} p}\)
    \(=\frac{\sin ^{5} p+\cos ^{5} p}{\sin ^{2} p \cos ^{2} p}=\frac{\sin ^{3} p\left(1-\cos ^{2} p\right)+\cos ^{3} p\left(1-\sin ^{2} p\right)}{\sin ^{2} p \cos ^{2} p}\)
    \(=\frac{\sin ^{3} p-\sin ^{3} p \cos ^{2} p+\cos ^{3} p-\cos ^{3} p \sin ^{2} p}{\sin ^{2} p \cos ^{2} p}\)
    \(=\frac{\sin ^{3} p+\cos ^{3} p-\sin ^{2} p \cos ^{2} p(\sin p+\cos p)}{\sin ^{2} p \cos ^{2} p}\)
    \(=\frac{(\sin p+\cos p)^{3}-3 \sin p \cos p(\sin p+\cos p)-\sin ^{2} p \cos ^{2} p(\sin p+\cos p)}{\sin ^{2} p \cos ^{2} p}\)
    \(=\frac{\left(\frac{1}{2}\right)^{3}-3\left(-\frac{3}{8}\right)\left(\frac{1}{2}\right)-\left(\frac{9}{64}\right)\left(\frac{1}{2}\right)}{\left(\frac{9}{64}\right)}=\frac{\frac{1}{8}+\frac{9}{16}-\frac{9}{128}}{\frac{9}{64}}=\frac{79}{18}\)
    Hence, the correct option is C.
  • Question 7
    1 / -0
    \(\tan 20^{\circ}=k \Rightarrow \frac{\tan 250^{\circ}+\tan 340^{\circ}}{\tan 200^{\circ}-\tan 110^{\circ}}=\)
    Solution
    \(\tan 250^{\circ}=\tan \left(270^{\circ}-20^{\circ}\right)=\cot 20^{\circ}\)
    \(\tan 340^{\circ}=\tan \left(360^{\circ}-20^{\circ}\right)=-\tan 20^{\circ}\)
    \(\tan 200^{\circ}=\tan \left(180^{\circ}+20^{\circ}\right)=\tan 20^{\circ}\)
    \(\tan 110^{\circ}=\tan \left(90^{\circ}+20^{\circ}\right)=-\cot 20^{\circ}\)
    Using these values in the given expression we get, \(\frac{\tan 250^{\circ}+\tan 340^{\circ}}{\tan 200^{\circ}-\tan 110^{\circ}}=\frac{\cot 20^{\circ}-\tan 20^{\circ}}{\tan 20^{\circ}+\cot 20^{\circ}}=\frac{1-k^{2}}{1+k^{2}}\)
    Hence, the correct option is A.
  • Question 8
    1 / -0

    In the above figure, what is the perimeter of Δ ACD?

    Solution
    In \(\Delta A B D, \angle A B D+\angle A D B+\angle B A D=180^{\circ}\)
    \(\Rightarrow \angle B A D=60^{\circ}\)
    \(\therefore \angle B A C=60-15=45^{\circ}\)
    \(\therefore \angle A C B=45^{\circ}\) and \(\Delta A B C\) is an isosceles triangle. \(A B=B C=2 \sqrt{2}\)
    \(\tan \left(30^{\circ}\right)=\frac{A B}{B D}\)
    So, \(B D=2 \sqrt{6}\)
    \(C D=B D-B C=2 \sqrt{6}-2 \sqrt{2} \ldots(1)\)
    \(\sin \left(30^{\circ}\right)=\frac{A B}{A D}\)
    So, \(A D=4 \sqrt{2} \ldots(2)\)
    Now, using equations
    (1) and ( 2 ), we get the required perimeter to be \(A C+C D+A D=4+2 \sqrt{6}-2 \sqrt{2}+4 \sqrt{2}\)
    \(=4+2 \sqrt{6}+2 \sqrt{2}\)
    Hence, the correct option is C.
  • Question 9
    1 / -0

    cosec2Acot2A − sec2Atan2A − (cot2A − tan2A)(sec2A + cosec2A − 1) is equal to

    Solution
    \(\operatorname{cosec}^{2} A \cot ^{2} A-\sec ^{2} A \tan ^{2} A-\left(\cot ^{2} A-\tan ^{2} A\right)\left(\sec ^{2} A+\operatorname{cosec}^{2} A-1\right)\)
    \(=\operatorname{cosec}^{2} A \cot ^{2} A-\sec ^{2} A \tan ^{2} A-\cot ^{2} A \sec ^{2} A-\cot ^{2} A \operatorname{cosec}^{2} A+\cot ^{2} A\)
    \(+\tan ^{2} A \sec ^{2} A+\tan ^{2} A \operatorname{cosec} A-\tan ^{2} A\)
    \(=-\cot ^{2} A \sec ^{2} A+\cot ^{2} A+\tan ^{2} A \operatorname{cosec}^{2} A-\tan ^{2} A\)
    \(=-\cot ^{2} A\left(\sec ^{2} A-1\right)+\tan ^{2} A\left(\operatorname{cosec}^{2} A-1\right)\)
    \(=-\cot ^{2} A \tan ^{2} A+\tan ^{2} A \cot ^{2} A\)
    \(=-1+1\)
    \(=0\)
    Hence, the correct option is A.
  • Question 10
    1 / -0

    If an is an A.P and a1+a4+a7+...+a16=147, then a1+a6+a11+a16=

    Solution
    Given equation can be written as \(a_{1}+\left(a_{1}+3 d\right)+\left(a_{1}+6 d\right)+\ldots+a_{1}+15 d=147\)
    \(\Rightarrow 6 a_{1}+3 d(1+2+\ldots+5)=147\)
    \(\Rightarrow 2 a_{1}+15 d=49\)
    Now, \(a_{1}+a_{6}+a_{11}+a_{16}=4 a_{1}+5 d+10 d+15 d\)
    \(=4 a_{1}+30 d=2\left(2 a_{1}+15 d\right)=2 \times 49\)
    \(=98\)
    Hence, the correct option is B.
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