Prove that : (i) tan 36o + tan 9o + tan 36o tan 9o = 1 (ii) tan 13x – tan 9x – tan 4x = tan 13x tan 9x tan 4x
(i) \[tan\text{ }{{36}^{o}}~+\text{ }tan\text{ }{{9}^{o}}~+\text{ }tan\text{ }{{36}^{o}}~tan\text{ }{{9}^{o}}~=~1\] We know \[36{}^\circ \text{ }+\text{ }9{}^\circ \text{ }=\text{ }45{}^\circ \] So...
Prove that: (i) tan 8x – tan 6x – tan 2x = tan 8x tan 6x tan 2x (ii) tan π/12 + tan π/6 + tan π/12 tan π/6 = 1
\[\left( \mathbf{i} \right)tan\text{ }8x\text{ }-\text{ }tan\text{ }6x\text{ }-\text{ }tan\text{ }2x\text{ }=\text{}tan\text{}8x\text{}tan\text{ }6x\text{ }tan\text{ }2x\] Let us consider LHS:...
Prove that: (i) cos^2 A + cos^2 B – 2 cos A cos B cos (A +B) = sin^2 (A + B)
(ii) Solution: (i) \[co{{s}^{2}}A+co{{s}^{2}}B - 2cosAcosBcos\left(A+B \right)=si{{n}^{2}}\left(A+B\right)\] Let us consider LHS: \[co{{s}^{2}}A\text{ }+\text{...
Prove that: (i)sin2 B = sin2 A + sin^2 (A-B) – 2sin A cos B sin (A – B)
(ii) Solution: (i) \[si{{n}^{2}}B=si{{n}^{2}}A+si{{n}^{2}}\left( A-B \right)2sin\text{}A\text{}cosB\text{}sin\left(A\text{}-\text{}B\right)\] Let us consider RHS:...
Prove that:
(i) (ii) Solution: (i) \[=~tan\text{ }A\] \[=\text{ }RHS\] \[\therefore LHS\text{ }=\text{ }RHS\] Hence proved. (ii) \[=~tan\text{ }A\text{ }-\text{ }tan\text{...
prove that: (i) cos2 π/4 – sin2 π/12 = √3/4 (ii) sin2 (n + 1) A – sin2nA = sin (2n + 1) A sin A
(i) \[co{{s}^{2}}~\pi /4\text{ }-\text{ }si{{n}^{2}}~\pi /12\text{ }=\text{ }\surd 3/4\] Let us consider LHS: \[co{{s}^{2}}~\pi /4\text{ }-\text{ }si{{n}^{2}}~\pi /12\] Since, \[co{{s}^{2}}A\text{...
Prove that: (tan 69o + tan 66o) / (1 – tan 69o tan 66o) = -1
Let us consider LHS: \[(tan\text{ }{{69}^{o}}~+\text{ }tan\text{ }{{66}^{o}})\text{ }/\text{ }(1\text{ }\text{ }tan\text{ }{{69}^{o}}~tan\text{ }{{66}^{o}})\] Since, \[tan\text{ }\left( A\text{...
Prove that:
Solution: \[=\text{ }sin\text{ }{{90}^{o}}\] \[=\text{ }1\] \[=\text{ }RHS\] \[\therefore LHS\text{ }=\text{ }RHS\] Hence proved.
Prove that:
(i) (ii) Solution: (i) \[=\text{ }sin\text{ }{{90}^{o}}\] \[=\text{ }1\] \[=\text{ }RHS\] \[\therefore LHS\text{ }=\text{ }RHS\] Hence proved. (ii) \[=\text{ }sin\text{ }{{60}^{o}}\] \[=\text{...
Prove that: (cos 8degree – sin 8degree) / (cos 8degree + sin 8degree) = tan 37degree
\[(cos\text{ }{{8}^{o}}-\text{ }sin\text{ }{{8}^{o}})\text{ }/\text{ }(cos\text{ }{{8}^{o}}~+\text{ }sin\text{ }{{8}^{o}})\text{ }=\text{ }tan\text{ }{{37}^{o}}\] Let us consider LHS: \[(cos\text{...
Prove that: (i) (cos 11o + sin 11o) / (cos 11o – sin 11o) = tan 56o (ii) (cos 9o + sin 9o) / (cos 9o – sin 9o) = tan 54o
\[\left(\mathbf{i} \right)~(cos\text{}{{11}^{o}}~+\text{}sin\text{}{{11}^{o}})\text{}/\text{}(cos\text{}{{11}^{o}}-\text{}sin\text{}{{11}^{o}})\text{}=\text{}tan\text{}{{56}^{o}}\] Taking L.H.S:...
Prove that: (tan A + tan B) / (tan A – tan B) = sin (A + B) / sin (A – B)
Let us consider LHS: \[\left( tan\text{ }A\text{ }+\text{ }tan\text{ }B \right)\text{ }/\text{ }\left( tan\text{ }A\text{ }-\text{ }tan\text{}B \right)\] \[=\text{ }RHS\] \[\therefore LHS\text{...
Prove that: cos 7π/12 + cos π/12 = sin 5π/12 – sin π/12
Since, \[7\pi /12\text{ }=\text{ }105{}^\circ ,\text{ }\pi /12\text{ }=\text{ }15{}^\circ ;\text{ }5\pi /12\text{ }=\text{ }75{}^\circ \] Let us consider LHS: \[cos\text{ }105{}^\circ \text{...
If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following: tan (A + B)
\[cos\text{ }A\text{ }=\text{ }-12/13\text{ }and\text{ }cot\text{ }B\text{ }=\text{ }24/7\] Since, A lies in second quadrant, B in the third quadrant. Sine function is positive, in the second...
If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following: (i) sin (A + B) (ii) cos (A + B)
\[cos\text{ }A\text{ }=\text{ }-12/13\text{ }and\text{ }cot\text{ }B\text{ }=\text{ }24/7\] Since, A lies in second quadrant, B in the third quadrant. Sine function is positive, in the second...
Evaluate the following: (i) sin 360 cos 90 + cos 360 sin 90 (ii) cos 800 cos 200 + sin 800 sin 200
\[\left( \mathbf{i} \right)~sin\text{ }{{36}^{0}}~cos\text{ }{{9}^{0}}~+\text{ }cos\text{ }{{36}^{0}}~sin\text{ }{{9}^{0}}\] Since, \[sin\text{ }\left( A\text{ }+B \right)\text{ }=\text{ }sin\text{...
Evaluate the following: (i) sin 780 cos 180 – cos 780 sin 180 (ii) cos 470 cos 130 – sin 470 sin 130
\[\left( \mathbf{i} \right)~sin\text{ }{{78}^{0}}~cos\text{ }{{18}^{0}}-\text{ }cos\text{ }{{78}^{0}}~sin\text{ }{{18}^{0}}\] Since, \[sin\text{ }\left( A\text{ }-\text{ }B \right)\text{ }=\text{...
If sin A = 1/2, cos B = √3/2, where π/2
As per the question: \[Sin\text{ }A\text{ }=\text{ }1/2\text{ }and\text{ }cos\text{ }B\text{ }=\text{ }\surd 3/2,\] \[where\text{ }\pi /2\text{ }<A\text{ }<\text{ }\pi \text{ }and\text{...
If sin A = 1/2, cos B = 12/13, where π/2
As per the question given: \[sin\text{ }A\text{ }=\text{ }1/2,\text{ }cos\text{ }B\text{ }=\text{ }12/13,\] \[where\text{ }\pi /2<A\text{ }<\text{ }\pi \text{ }and\text{ }3\pi /2\text{...
If tan A = 3/4, cos B = 9/41, where π
As per the question given: \[tan\text{ }A\text{ }=\text{ }3/4\text{ }and\text{ }cos\text{ }B\text{ }=\text{ }9/41,\] \[where\text{ }\pi \text{ }<A\text{ }<\text{ }3\pi /2\text{ }and\text{...
If cos A = – 24/25 and cos B = 3/5, where π
As per the question given: \[cos\text{ }A\text{ }=-\text{ }\text{ }24/25\text{ }and\text{ }cos\text{ }B\text{ }=\text{ }3/5,\] where \[\pi \text{ }<A\text{ }<\text{ }3\pi /2\text{ }and\text{...
If sin A = 3/5, cos B = –12/13, where A and B, both lie in second quadrant, find the value of sin (A +B).
As per the question given: \[sin\text{ }A\text{ }=\text{ }3/5,\text{ }cos\text{ }B\text{ }=-\text{ }12/13,\] \[A\text{ }and\text{ }B,\]both lie in second quadrant. Since,...
If Sin A = 12/13 and sin B = 4/5, where π/2
As per the question given: \[Sin\text{ }A\text{ }=\text{ }12/13\text{ }and\text{ }sin\text{ }B\text{ }=\text{ }4/5,\] \[where\text{ }\pi /2<A\text{ }<\text{ }\pi \text{ }and\text{ }0\text{...
If sin A = 4/5 and cos B = 5/13, where 0
As per the question given: \[sin\text{ }A\text{ }=\text{ }4/5\text{ }and\text{ }cos\text{ }B\text{ }=\text{ }5/13\]...
If sin A = 4/5 and cos B = 5/13, where 0
As per the question given: \[sin\text{ }A\text{ }=\text{ }4/5\text{ }and\text{ }cos\text{ }B\text{ }=\text{ }5/13\] Since, \[cos\text{ }A\text{}=\text{}\surd...