(i) sin 60° cos 30° + sin 30° cos 60°
(ii) $2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°$
(iii)$\dfrac{cos 45^\circ}{sec 30^\circ+cosec 30^\circ}$
(iv)$\dfrac{sin 30^\circ+tan 45^\circ-cosec 60^\circ}{sec 30^\circ+cos60^\circ+cot45^\circ}$
(v)$\dfrac{5cos^{2}60^\circ+4sec^{2}30^\circ-tan^{2}45^\circ}{sin^{2}30^\circ+cos^{2}30^\circ}$
(i) sin 60° cos 30° + sin 30° cos 60°
First, find the values of the given trigonometric ratios
sin 30° = $\dfrac{1}{2}$
cos 30° = $\dfrac{\sqrt{3}}{2}$
sin 60° = $\dfrac{3}{2}$
cos 60°= $\dfrac{1}{2}$
Now, substitute the values in the given problem
sin 60° cos 30° + sin 30° cos 60° = $\dfrac{\sqrt{3}}{2}×\dfrac{\sqrt{3}}{2} + (\dfrac{1}{2}) ×(\dfrac{1}{2} )$ = $\dfrac{3}{4}+\dfrac{1}{4}$ = $\dfrac{4}{4}$ =1
(ii) $2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°$
We know that, the values of the trigonometric ratios are:
sin 60° =$\dfrac{ \sqrt{3}}{2}$
cos 30° = $\dfrac{\sqrt{3}}{2}$
tan 45° = 1
Substitute the values in the given problem
$2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°$ = $2(1)^{2} + (\sqrt{3}/2)^{2}-(\sqrt{3}/2)^{2}$
$2 tan^{2}45° + cos^{2} 30° – sin^{2} 60° $= 2 + 0
$2 tan^{2} 45° + cos^{2} 30° – sin^{2} 60°$ = 2
(iii) $\dfrac{cos 45°}{(sec 30°+cosec 30°)}$
We know that,
cos 45° = $\dfrac{1}{\sqrt{2}}$
sec 30° = $\dfrac{2}{\sqrt{3}}$
cosec 30° = 2
Substitute the values, we get
$\dfrac{cos 45^\circ}{sec 30^\circ+cosec 30^\circ}$
=$\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}$
=$\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}$
=$\dfrac{1}{\sqrt{2}}$ x $\dfrac{\sqrt{3}}{2+2\sqrt{3}}$
=$\dfrac{1}{\sqrt{2}}$ x $\dfrac{\sqrt{3}}{2(1+\sqrt{3}}$
=$\dfrac{\sqrt{3}}{2\sqrt{2}(1+\sqrt{3})}$
=$\dfrac{\sqrt{3}}{2\sqrt{2}(\sqrt{3}+1)}$
Now,rationalize the terms we get,
=$\dfrac{\sqrt{2}}{2\sqrt{2}(\sqrt{3}+1)}$x$\dfrac{\sqrt{3}-1}{\sqrt{3}-1}$
=$\dfrac{3-\sqrt{3}}{2\sqrt{2}(3-1)}$
=$\dfrac{3-\sqrt{3}}{2\sqrt{2}(2)}$
Now, multiply both the numerator and denominator by $\sqrt{2}$ , we get
=$\dfrac{3-\sqrt{3}}{2\sqrt{2}(2)}$x$\dfrac{\sqrt{2}}{\sqrt{2}}$
=$\dfrac{3\sqrt{2}-\sqrt{3}\sqrt{2}}{8}$
=$\dfrac{3\sqrt{2}-\sqrt{6}}{8}$
Therefore, $\dfrac{cos 45°}{(sec 30°+cosec 30°)}$ = $\dfrac{(3\sqrt{2} – \sqrt{6})}{8}$
(iv)$\dfrac{sin 30^\circ+tan 45^\circ-cosec 60^\circ}{sec 30^\circ+cos60^\circ+cot45^\circ}$
We know that,
sin 30° = $\dfrac{1}{2}$
tan 45° = 1
cosec 60° = $\dfrac{2}{\sqrt{3}}$
sec 30° = $\dfrac{2}{\sqrt{3}}$
cos 60° = $\dfrac{1}{2}$
cot 45° = 1
Substitute the values in the given problem, we get
$\dfrac{sin 30^\circ+tan 45^\circ-cosec 60^\circ}{sec 30^\circ+cos60^\circ+cot45^\circ}$
=$\dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$
=$\dfrac{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}$
Now cancel the term $ 2\sqrt{3}$,in numerator and denominator ,we get
=$\dfrac{\sqrt{3}+2\sqrt{3}-4}{4+\sqrt{3}+2\sqrt{3}}$
=$\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$
Now,rationalize the terms
=$\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$x$\dfrac{3\sqrt{3}}{3\sqrt{3}+4}$
=$\dfrac{27-12\sqrt{3}-12\sqrt{3}+16}{27-12\sqrt{3}+12\sqrt{3}+16}$
=$\dfrac{27-24\sqrt{3}+16}{11}$
=$\dfrac{43-24\sqrt{3}}{11}$
Therefore,
$\dfrac{sin 30^\circ+tan 45^\circ-cosec 60^\circ}{sec 30^\circ+cos60^\circ+cot45^\circ}$=$\dfrac{43-24\sqrt{3}}{11}$
(v)$\dfrac{5cos^{2}60^\circ+4sec^{2}30^\circ-tan^{2}45^\circ}{sin^{2}30^\circ+cos^{2}30^\circ}$
We know that,
cos 60° = $\dfrac{1}{2}$
sec 30° = $\dfrac{ 2}{\sqrt{3}}$
tan 45° = 1
sin 30° = $\dfrac{1}{2}$
cos 30° = $\dfrac{\sqrt{3}}{2}$
Now, substitute the values in the given problem, we get
$(5cos^{2}60° + 4sec^{2}30° – \dfrac{tan^{2}45°)}{(sin^{2} 30°} + cos^{2} 30°)$
= $\dfrac{5(\dfrac{1}{2})^{2}+4(\dfrac{2}{\sqrt{3}})^{2}-1^{2}}{(\dfrac{1}{2})^{2}+(\dfrac{\sqrt{3}}{2})^{2}}$
= $\dfrac{(\dfrac{5}{4}+\dfrac{16}{3}-1)}{(\dfrac{1}{4}+\dfrac{3}{4})}$
= $\dfrac{\dfrac{(15+64-12)}{12}}{(\dfrac{4}{4})}$
=$ \dfrac{67}{12}$
(i) $\dfrac{2tan 30°}{1+tan^{2}30°}$ =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) $\dfrac{1-tan^{2}45°}{1+tan^{2}45°}$ =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) $\dfrac{2tan30°}{1-tan^{2}30°}$ =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
(i) (A) is correct.
Substitute the value of tan 30° in the given equation
tan 30° = $\dfrac{1}{\sqrt{3}}$
$\dfrac{2tan 30°}{1+tan^{2}30°}$ = $\dfrac{2(\dfrac{1}{\sqrt{3}})}{1+(\dfrac{1}{\sqrt{3}})^{2}}$
= $\dfrac{(\dfrac{2}{\sqrt{3}})}{(1+\dfrac{1}{3})}$ = $\dfrac{(\dfrac{2}{\sqrt{3}})}{(\dfrac{4}{3})}$
= $\dfrac{6}{4\sqrt{3}}$ = $\dfrac{\sqrt{3}}{2}$ = sin 60°
The obtained solution is equivalent to the trigonometric ratio sin 60°
(ii) (D) is correct.
Substitute the value of tan 45° in the given equation
tan 45° = 1
$\dfrac{1-tan^{2}45°}{1+tan^{2}45°}$ = $\dfrac{(1-1^{2})}{(1+1^{2})}$
= $\dfrac{0}{2}$ = 0
The solution of the above equation is 0.
(iii) (A) is correct.
To find the value of A, substitute the degree given in the options one by one
sin 2A = 2 sin A is true when A = 0°
As sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2 × 0 = 0
or,
Apply the sin 2A formula, to find the degree value
sin 2A = 2sin A cos A
⇒2sin A cos A = 2 sin A
⇒ 2cos A = 2 ⇒ cos A = 1
Now, we have to check, to get the solution as 1, which degree value has to be applied.
When 0 degree is applied to cos value, i.e., cos 0 =1
Therefore, ⇒ A = 0°
(iv) (C) is correct.
Substitute the value of tan 30° in the given equation
tan 30° = $\dfrac{1}{\sqrt{3}}$
$\dfrac{2tan30°}{1-tan^{2}30°}$ = $\dfrac{2(\dfrac{1}{\sqrt{3}})}{1-(\dfrac{1}{\sqrt{3}})^{2}}$
=$\dfrac{ (\dfrac{2}{\sqrt{3}})}{(1-\dfrac{1}{3})}$ = $\dfrac{(\dfrac{2}{\sqrt{3}})}{(\dfrac{2}{3})}$ = $\sqrt{3}$ = tan 60°
The value of the given equation is equivalent to tan 60°.
tan (A + B) = $\sqrt{3}$
Since $\sqrt{3}$ = tan 60°
Now substitute the degree value
⇒ tan (A + B) = tan 60°
(A + B) = 60° … (i)
The above equation is assumed as equation (i)
tan (A – B) = $\dfrac{1}{\sqrt{3}}$
Since $\dfrac{1}{\sqrt{3}}$ = tan 30°
Now substitute the degree value
⇒ tan (A – B) = tan 30°
(A – B) = 30° … equation (ii)
Now add the equation (i) and (ii), we get
A + B + A – B = 60° + 30°
Cancel the terms B
2A = 90°
A= 45°
Now, substitute the value of A in equation (i) to find the value of B
45° + B = 60°
B = 60° – 45°
B = 15°
Therefore A = 45° and B = 15°
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
(i) False.
Justification:
Let us take A = 30° and B = 60°, then
Substitute the values in the sin (A + B) formula, we get
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°
= $\dfrac{1}{2}$ + $\dfrac{\sqrt{3}}{2}$ = 1+$\dfrac{\sqrt{3}}{2}$
Since the values obtained are not equal, the solution is false.
(ii) True.
Justification:
According to the values obtained as per the unit circle, the values of sin are:
sin 0° = 0
sin 30° =$\dfrac{ 1}{2}$
sin 45° = $\dfrac{1}{\sqrt{2}}$
sin 60° = $\dfrac{\sqrt{3}}{2}$
sin 90° = 1
Thus the value of sin θ increases as θ increases. Hence, the statement is true
(iii) False.
According to the values obtained as per the unit circle, the values of cos are:
cos 0° = 1
cos 30° =$\dfrac{ \sqrt{3}}{2}$
cos 45° = $\dfrac{1}{\sqrt{2}}$
cos 60° = 1/2
cos 90° = 0
Thus, the value of cos θ decreases as θ increases. So, the statement given above is false.
(iv) False
sin θ = cos θ, when a right triangle has 2 angles of ($\dfrac{π}{4}$). Therefore, the above statement is false.
(v) True.
Since cot function is the reciprocal of the tan function, it is also written as:
cot A = $\dfrac{cos A}{sin A}$
Now substitute A = 0°
cot 0° = $\dfrac{cos 0°}{sin 0°}$ = $\dfrac{1}{0}$ = undefined.
Hence, it is true