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Excercise 15B
Prove that:
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Prove that:
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Prove that:
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If $\cos x=\frac{3}{5}\,and\,\cos y=-\frac{24}{25}$, where $\frac{3\pi }{2}
Answer: (iii)Here first we will calculate value of tanx and tany,
If $\cos x=\frac{3}{5}\,and\,\cos y=-\frac{24}{25}$, where $\frac{3\pi }{2}
Answer: Given: $\cos x=\frac{3}{5}\,and\,\cos y=-\frac{24}{25}$ We will first find out value of sinx and siny, (i)sin(x + y) = sinx.cosy + cosx.siny (ii)cos(x - y) = cosx.cosy +...
If $\sin x=\frac{12}{13}\,and\,\sin y=\frac{4}{5}$, where $\frac{\pi }{2}
Answer: Given: $\sin x=\frac{12}{13}\,and\,\sin y=\frac{4}{5}$ (iii)Here first we will calculate value of tanx and tany
If $\sin x=\frac{12}{13}\,and\,\sin y=\frac{4}{5}$, where $\frac{\pi }{2}
Answer: Given: $\sin x=\frac{12}{13}\,and\,\sin y=\frac{4}{5}$ Here we will find values of cosx and cosy (i) sin(x + y) = sinx.cosy + cosx.siny (ii)cos(x + y) = cosx.cosy +...
If x and y are acute angles such that $\cos x=\frac{13}{14}\,and\,\cos y=\frac{1}{7}$, prove that $\left( x-y \right)=-\frac{\pi }{3}$
Answer: Given: $\cos x=\frac{13}{14}\,and\,\cos y=\frac{1}{7}$ Now we will calculate value of sinx and siny Hence, Cos(x - y) = cosx.cosy + sinx.siny
If x and y are acute such that $\sin x=\frac{1}{\sqrt{5}}\,and\,\sin y=\frac{1}{\sqrt{10}}$, prove that $(x+y)=\frac{\pi }{4}$
Answer: Given: $\sin x=\frac{1}{\sqrt{5}}\,and\,\sin y=\frac{1}{\sqrt{10}}$ Now we will calculate value of cos x and cosy Sin(x + y) = sinx.cosy + cosx.siny
If θ and Φ lie in the first quadrant such that $\sin \theta =\frac{8}{17}\,and\,\cos \phi =\frac{12}{13}$, find the values of (iii) tan (θ – Φ)
Answer: (iii)We will first find out the Values of tanθ and tanΦ
If θ and Φ lie in the first quadrant such that $\sin \theta =\frac{8}{17}\,and\,\cos \phi =\frac{12}{13}$, find the values of (i) sin (θ – Φ ) (ii) cos (θ – Φ)
Answer: Given: $\sin \theta =\frac{8}{17}\,and\,\cos \phi =\frac{12}{13}$
Prove that:
Answer: Using cos(90° + θ) = - sinθ(I quadrant cosx is positive cosec( - θ) = - cosecθ tan(270° - θ) = tan(180° + 90° - θ) = tan(90° - θ) = cotθ (III quadrant tanx is positive) Similarily sin(270° +...
Prove that:
Answer: Using sin(90° + θ) = cosθ and sin( - θ) = sinθ,tan(90° + θ) = - cotθ Sin(180° + θ) = - sinθ(III quadrant sinx is negative)
Prove that
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Prove that:
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Prove that:
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Prove that:
Answer: (II quadrant tanx negative) - tan45° = - 1
Prove that
Answer: (ii)cot105° - tan105° = cot(180° - 75°) - tan(180° - 75°) (II quadrant tanx is negative and cotx as well) = - cot75° - ( - tan75°) = tan75° - cot75°
Prove that: tan15° + cot15° = 4
Answer: (iii) tan15° + cot15° = First, we will calculate tan15°,
Prove that:
Answer: (i) sin75° = sin(90° - 15°) .…….(using sin(A - B) = sinAcosB - cosAsinB) = sin90°cos15° - cos90°sin15° = 1.cos15° - 0.sin15° = cos15° Cos15° = cos(45° - 30°) …………(using cos(A - B) = cosAcosB...
Prove that:
Answer:
Prove that:
(i) cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x = cos x Answer: (i) cos(n + 2)x.cos(n + 1)x + sin(n + 2)x.sin(n + 1)x = sin((n + 2)x + (n + 1)x)(using cos(A - B) = cosAcosB + sinAsinB) =...
Prove that (i) sin(50° + θ)cos(20° + θ) – cos(50° + θ)sin(20° + θ) = 1/2 (ii) cos(70° + θ)cos(10° + θ) + sin(70° + θ)sin(10° + θ) = 1/2
Answer: (i) We have: sin(50° + θ)cos(20° + θ) - cos(50° + θ)sin(20° + θ) = sin(50° + θ - (20° + θ))(using sin(A - B) = sinAcosB - cosAsinB) = sin(50° + θ - 20° - θ) = sin30° = 1/2 (ii) We have:...
Prove that (v) cos130°cos40° + sin130°sin40° = 0
(v) cos130°cos40° + sin130°sin40° = cos(130° - 40°) (using cos(A - B) = cosAcosB + sinAsinB) = cos90° = 0
Prove that
Answer: (iii) cos75°cos15° + sin75°sin15° = cos(75° - 15°) (using cos(A - B) = cosAcosB + sinAsinB) = cos60° = 1/2 (iv) sin40°cos20° + cos40°sin20° = sin(40° + 20°) (using sin(A + B) = sinAcosB +...
Prove that
Answer: (i) sin80°cos20° - cos80°sin20° = sin(80° - 20°) (using sin(A - B) = sinAcosB - cosAsinB) = sin60° = $\frac{\sqrt{3}}{2}$ (ii)cos45°cos15° - sin45°sin15° = cos(45° + 15°) (using cos(A + B) =...
Find the values of all trigonometric functions of 135 deg
Answer:
Find the value of (ix) cos (495֯ )
(ix)cos495° = cos(360° + 135°) …………(using cos(360° + x) = cosx) = cos135° = cos(180° - 45°) ………….(using cos(180° - x) = - cosx) = - cos45° = - 1//2
Find the value of (vii) cot ( – 315֯ ) (viii) sin ( – 1230֯ )
Answer: vii) $ \cot \left( -{{315}^{\circ }} \right)=\frac{1}{\tan \left( -{{315}^{\circ }} \right)} $ $ \Rightarrow \frac{1}{-\tan \left( {{315}^{\circ }} \right)}=\frac{1}{-\tan \left(...
Find the value of (v) cosec ( – 690֯ ) (vi) tan (225֯ )
Answer:
Find the value of (iii) tan ( – 120֯ ) (iv) sec ( – 420֯ )
Answer: (iii) tan( - 120°) = - tan12 …….(tan( - x) = tanx) = - tan(180° - 60°) ……. (in II quadrant tanx is negative) = - ( - tan60°) = tan60°