Mix Examples - Introduction to Trigonometry Questions in English

(D) Given $\sin \theta = \frac{3}{5}$.
We know that $\cos^2 \theta = 1 - \sin^2 \theta$.
$\cos^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$.
Therefore,$\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$.

(C) Given that $\sec \theta = \frac{5}{3}$.
We use the trigonometric identity $\tan^2 \theta = \sec^2 \theta - 1$.
Substituting the value of $\sec \theta$:
$\tan^2 \theta = \left(\frac{5}{3}\right)^2 - 1$
$\tan^2 \theta = \frac{25}{9} - 1$
$\tan^2 \theta = \frac{25 - 9}{9} = \frac{16}{9}$
Taking the square root on both sides:
$\tan \theta = \sqrt{\frac{16}{9}} = \frac{4}{3}$.

(A) We know that $\frac{1}{\cos ^{2} \theta} = \sec ^{2} \theta$ and $\frac{1}{\cot ^{2} \theta} = \tan ^{2} \theta$.
Substituting these into the expression,we get $\sec ^{2} \theta - \tan ^{2} \theta$.
Using the trigonometric identity $1 + \tan ^{2} \theta = \sec ^{2} \theta$,we have $\sec ^{2} \theta - \tan ^{2} \theta = 1$.

(B) We know the trigonometric identities:
$1 + \tan^{2} \theta = \sec^{2} \theta$ and $1 - \cos^{2} \theta = \sin^{2} \theta$.
Substituting these into the expression:
$(1 + \tan^{2} \theta)(1 - \cos^{2} \theta) = \sec^{2} \theta \cdot \sin^{2} \theta$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we have:
$\sec^{2} \theta \cdot \sin^{2} \theta = \frac{1}{\cos^{2} \theta} \cdot \sin^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta} = \tan^{2} \theta$.

(A) Given equations are:
$a \sin \theta = 3$ --- $(1)$
$a \cos \theta = 4$ --- $(2)$
Squaring both equations:
$(a \sin \theta)^2 = 3^2 \implies a^2 \sin^2 \theta = 9$ --- $(3)$
$(a \cos \theta)^2 = 4^2 \implies a^2 \cos^2 \theta = 16$ --- $(4)$
Adding equations $(3)$ and $(4)$:
$a^2 \sin^2 \theta + a^2 \cos^2 \theta = 9 + 16$
$a^2 (\sin^2 \theta + \cos^2 \theta) = 25$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$a^2 (1) = 25$
$a^2 = 25$
Since $a > 0$,we take the positive square root:
$a = 5$.

(B) Given equation: $\sin 5 \theta = \cos 5 \theta$
Divide both sides by $\cos 5 \theta$ (assuming $\cos 5 \theta \neq 0$):
$\frac{\sin 5 \theta}{\cos 5 \theta} = 1$
$\tan 5 \theta = 1$
Since $\tan 45^{\circ} = 1$,we have:
$5 \theta = 45^{\circ}$
$\theta = \frac{45^{\circ}}{5} = 9^{\circ}$
Thus,the value of $\theta$ is $9$.

(C) Given $\cos \theta = \frac{15}{17}$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we have:
$\sin^2 \theta = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289}$.
Therefore,$\sin \theta = \sqrt{\frac{64}{289}} = \frac{8}{17}$.
Now,$\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{17}{8}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{15/17}{8/17} = \frac{15}{8}$.
Finally,$\operatorname{cosec} \theta + \cot \theta = \frac{17}{8} + \frac{15}{8} = \frac{32}{8} = 4$.

(D) We know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$.
Substituting these into the expression: $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$.
Taking the common denominator: $\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$.
Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$,we get: $\frac{1}{\sin \theta \cos \theta}$.
Since $\frac{1}{\sin \theta} = \operatorname{cosec} \theta$ and $\frac{1}{\cos \theta} = \sec \theta$,the expression simplifies to $\operatorname{cosec} \theta \cdot \sec \theta$.

(B) We are given the expression: $(\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2$.
Using the algebraic identities $(a+b)^2 = a^2 + 2ab + b^2$ and $(a-b)^2 = a^2 - 2ab + b^2$,we expand the terms:
$= (\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta) + (\sin^2 \theta - 2 \sin \theta \cos \theta + \cos^2 \theta)$.
Combining like terms,the $2 \sin \theta \cos \theta$ and $-2 \sin \theta \cos \theta$ cancel out:
$= \sin^2 \theta + \cos^2 \theta + \sin^2 \theta + \cos^2 \theta$.
$= 2 \sin^2 \theta + 2 \cos^2 \theta$.
$= 2(\sin^2 \theta + \cos^2 \theta)$.
Since the fundamental trigonometric identity is $\sin^2 \theta + \cos^2 \theta = 1$,we substitute this value:
$= 2(1) = 2$.

(C) We know the identity $\sec^{2} \theta = 1 + \tan^{2} \theta$.
Substituting the given value: $\sec^{2} \theta = 1 + (\frac{5}{12})^{2}$.
$\sec^{2} \theta = 1 + \frac{25}{144} = \frac{144 + 25}{144} = \frac{169}{144}$.
Taking the square root,$\sec \theta = \frac{13}{12}$.
Since $\cos \theta = \frac{1}{\sec \theta}$,we get $\cos \theta = \frac{12}{13}$.

(C) Given that $A = 30^{\circ}$.
We need to find the value of $\cos 2A$.
Substitute the value of $A$ into the expression:
$\cos 2A = \cos(2 \times 30^{\circ})$
$\cos 2A = \cos 60^{\circ}$
Since $\cos 60^{\circ} = \frac{1}{2}$,the value is $\frac{1}{2}$.

(B) We know the trigonometric identity for complementary angles: $\tan \theta = \cot(90^{\circ} - \theta)$.
Substituting $\theta = 15^{\circ}$ into the identity,we get:
$\tan 15^{\circ} = \cot(90^{\circ} - 15^{\circ})$.
$\tan 15^{\circ} = \cot 75^{\circ}$.
Therefore,the value of $\tan 15^{\circ}$ is equal to $\cot 75^{\circ}$.

(A) We know the fundamental trigonometric identity: $\sin ^{2} A + \cos ^{2} A = 1$.
Comparing the given equation $\sin ^{2} 35^{\circ} + \cos ^{2} \theta = 1$ with the identity $\sin ^{2} A + \cos ^{2} A = 1$,we can see that the angles must be equal for the sum to be $1$.
Therefore,$\theta = 35^{\circ}$.

(B) We know that the trigonometric identity for complementary angles is $\cos (90^\circ - \theta) = \sin \theta$.
Substituting this into the given expression:
$\sin \theta \cdot \cos (90^\circ - \theta) = \sin \theta \cdot \sin \theta$.
Therefore,the expression simplifies to $\sin^2 \theta$.

(B) Given the equation: $\tan ^{2} \theta = \sin ^{2} \theta + \cos ^{2} \theta$
We know the fundamental trigonometric identity: $\sin ^{2} \theta + \cos ^{2} \theta = 1$
Substituting this identity into the equation,we get: $\tan ^{2} \theta = 1$
Taking the square root on both sides,and considering $0 < \theta < 90^{\circ}$ (where $\tan \theta$ is positive): $\tan \theta = 1$
Since $\tan 45^{\circ} = 1$,we conclude that $\theta = 45^{\circ}$.

(B) Given the equation $\sin ^{2}(3 x+30^{\circ})+\cos ^{2}(2 x+45^{\circ})=1$.
We know the fundamental trigonometric identity $\sin ^{2} \theta+\cos ^{2} \theta=1$.
Comparing the given equation with the identity,we have $3 x+30^{\circ} = 2 x+45^{\circ}$.
Subtracting $2x$ from both sides,we get $x+30^{\circ} = 45^{\circ}$.
Subtracting $30^{\circ}$ from both sides,we get $x = 15^{\circ}$.

(A) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\tan 45^{\circ} = 1$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 90^{\circ} = 0$.
Substituting these values into the expression:
$\sin^{2} 60^{\circ} - \tan 45^{\circ} + \cos^{2} 30^{\circ} - \cot 90^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^{2} - 1 + \left(\frac{\sqrt{3}}{2}\right)^{2} - 0$
$= \frac{3}{4} - 1 + \frac{3}{4} - 0$
$= \frac{3}{4} + \frac{3}{4} - 1$
$= \frac{6}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.

(D) We know the trigonometric identities for complementary angles: $\tan(90^\circ - A) = \cot A$ and $\sec(90^\circ - A) = \operatorname{cosec} A$.
First,consider the term $\tan(65^\circ - \theta)$:
$\tan(65^\circ - \theta) = \cot(90^\circ - (65^\circ - \theta)) = \cot(25^\circ + \theta)$.
Next,consider the term $\sec(55^\circ - \theta)$:
$\sec(55^\circ - \theta) = \operatorname{cosec}(90^\circ - (55^\circ - \theta)) = \operatorname{cosec}(35^\circ + \theta)$.
Now,substitute these into the given expression:
$\tan(65^\circ - \theta) - \cot(25^\circ + \theta) - \sec(55^\circ - \theta) + \operatorname{cosec}(35^\circ + \theta)$
$= \cot(25^\circ + \theta) - \cot(25^\circ + \theta) - \operatorname{cosec}(35^\circ + \theta) + \operatorname{cosec}(35^\circ + \theta)$
$= 0$.

(B) Given expression: $8 \sin^{2} 45^{\circ} - 2 \tan^{2} 60^{\circ} + 3 \cot^{2} 30^{\circ} - 2 \cos^{2} 45^{\circ}$
Substitute the trigonometric values:
$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\tan 60^{\circ} = \sqrt{3}$,$\cot 30^{\circ} = \sqrt{3}$,$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$
$= 8 \left( \frac{1}{\sqrt{2}} \right)^{2} - 2 (\sqrt{3})^{2} + 3 (\sqrt{3})^{2} - 2 \left( \frac{1}{\sqrt{2}} \right)^{2}$
$= 8 \left( \frac{1}{2} \right) - 2 (3) + 3 (3) - 2 \left( \frac{1}{2} \right)$
$= 4 - 6 + 9 - 1$
$= 6$

(C) We know that $\frac{1}{\cos \theta} = \sec \theta$.
Therefore,$\frac{1}{\cos ^{2} \theta} = \sec ^{2} \theta$.
Using the trigonometric identity $1 + \tan ^{2} \theta = \sec ^{2} \theta$,we can write $\sec ^{2} \theta - 1 = \tan ^{2} \theta$.
Thus,$\frac{1}{\cos ^{2} \theta} - 1 = \tan ^{2} \theta$.

(B) We know that $\frac{1}{\sin \theta} = \operatorname{cosec} \theta$.
Therefore,$\frac{1}{\sin ^{2} \theta} = \operatorname{cosec}^{2} \theta$.
Substituting this into the expression: $\operatorname{cosec}^{2} \theta - 1$.
Using the trigonometric identity $1 + \cot^{2} \theta = \operatorname{cosec}^{2} \theta$,we get $\operatorname{cosec}^{2} \theta - 1 = \cot^{2} \theta$.

(A) We know the fundamental trigonometric identity: $\sec ^{2} \theta - \tan ^{2} \theta = 1$.
To find the value of $\tan ^{2} \theta - \sec ^{2} \theta$,we can multiply the identity by $-1$.
$-1 \times (\sec ^{2} \theta - \tan ^{2} \theta) = -1 \times 1$.
Therefore,$\tan ^{2} \theta - \sec ^{2} \theta = -1$.

(B) We know the algebraic identity $a^{2}-b^{2}=(a-b)(a+b)$.
Applying this to the expression $\sec ^{4} \theta-\tan ^{4} \theta$,we get:
$\sec ^{4} \theta-\tan ^{4} \theta = (\sec ^{2} \theta - \tan ^{2} \theta)(\sec ^{2} \theta + \tan ^{2} \theta)$.
From the fundamental trigonometric identity,we know that $\sec ^{2} \theta - \tan ^{2} \theta = 1$.
Substituting the given values:
$\sec ^{4} \theta - \tan ^{4} \theta = (1) \times \left(\frac{13}{12}\right) = \frac{13}{12}$.

(C) Given that $\tan \theta = 1$.
Since $\tan 45^{\circ} = 1$,we have $\theta = 45^{\circ}$.
Now,substitute $\theta = 45^{\circ}$ into the expression $\sin \theta \cdot \cos \theta$:
$\sin 45^{\circ} \cdot \cos 45^{\circ} = \left( \frac{1}{\sqrt{2}} \right) \cdot \left( \frac{1}{\sqrt{2}} \right) = \frac{1}{2}$.

(D) We are given the expression $\frac{\sec \theta-1}{\sec \theta+1}$.
Since $\sec \theta = \frac{1}{\cos \theta}$,we can multiply the numerator and the denominator by $\cos \theta$:
$\frac{\sec \theta-1}{\sec \theta+1} = \frac{(\sec \theta-1) \cdot \cos \theta}{(\sec \theta+1) \cdot \cos \theta}$
$= \frac{\sec \theta \cdot \cos \theta - \cos \theta}{\sec \theta \cdot \cos \theta + \cos \theta}$
$= \frac{1 - \cos \theta}{1 + \cos \theta}$ (since $\sec \theta \cdot \cos \theta = 1$)
Thus,the correct option is $D$.

(D) We know that $\cot \theta = \frac{1}{\tan \theta}$.
Substituting this into the expression,we get:
$\cot \theta \cdot \tan \theta = \left( \frac{1}{\tan \theta} \right) \cdot \tan \theta = 1$.

(C) Given equation: $\cos \theta = \sin 5 \theta$
We know that $\sin(90^{\circ} - A) = \cos A$,so $\sin 5 \theta = \cos(90^{\circ} - 5 \theta)$.
Substituting this into the equation: $\cos \theta = \cos(90^{\circ} - 5 \theta)$.
Since the cosine function is one-to-one for acute angles,we can equate the arguments:
$\theta = 90^{\circ} - 5 \theta$
$\theta + 5 \theta = 90^{\circ}$
$6 \theta = 90^{\circ}$
$\theta = \frac{90^{\circ}}{6} = 15^{\circ}$.

(A) Given expression: $\sin^{2} 30^{\circ} - \tan 45^{\circ} + \cos^{2} 60^{\circ} - \cot 90^{\circ}$
Substitute the trigonometric values:
$\sin 30^{\circ} = \frac{1}{2}$,$\tan 45^{\circ} = 1$,$\cos 60^{\circ} = \frac{1}{2}$,and $\cot 90^{\circ} = 0$
Substituting these values into the expression:
$= (\frac{1}{2})^{2} - 1 + (\frac{1}{2})^{2} - 0$
$= \frac{1}{4} - 1 + \frac{1}{4}$
$= \frac{2}{4} - 1$
$= \frac{1}{2} - 1$
$= -\frac{1}{2}$

(D) Given equation: $\tan 7 \theta \cdot \tan 3 \theta = 1$
We know that $\tan 3 \theta = \frac{1}{\cot 3 \theta}$,so $\tan 7 \theta = \frac{1}{\tan 3 \theta} = \cot 3 \theta$.
Using the trigonometric identity $\cot \alpha = \tan(90^{\circ} - \alpha)$,we can write:
$\tan 7 \theta = \tan(90^{\circ} - 3 \theta)$
Equating the angles:
$7 \theta = 90^{\circ} - 3 \theta$
Adding $3 \theta$ to both sides:
$10 \theta = 90^{\circ}$
Dividing by $10$:
$\theta = 9^{\circ}$

(C) Given expression: $\tan 5^{\circ} \cdot \tan 25^{\circ} \cdot \tan 45^{\circ} \cdot \tan 65^{\circ} \cdot \tan 85^{\circ}$
We know that $\tan 45^{\circ} = 1$ and $\tan(90^{\circ} - \theta) = \cot \theta$.
Substituting these values:
$= \tan 5^{\circ} \cdot \tan 25^{\circ} \cdot 1 \cdot \tan(90^{\circ} - 25^{\circ}) \cdot \tan(90^{\circ} - 5^{\circ})$
$= \tan 5^{\circ} \cdot \tan 25^{\circ} \cdot 1 \cdot \cot 25^{\circ} \cdot \cot 5^{\circ}$
Since $\tan \theta \cdot \cot \theta = 1$,we group the terms:
$= (\tan 5^{\circ} \cdot \cot 5^{\circ}) \cdot (\tan 25^{\circ} \cdot \cot 25^{\circ}) \cdot 1$
$= 1 \cdot 1 \cdot 1 = 1$

(A) We know that $\sin(90^{\circ} - \theta) = \cos \theta$.
Therefore,$\sin 75^{\circ} = \sin(90^{\circ} - 15^{\circ}) = \cos 15^{\circ}$.
Substituting this into the expression:
$\sin^{2} 15^{\circ} + \sin^{2} 75^{\circ} = \sin^{2} 15^{\circ} + (\cos 15^{\circ})^{2}$
$= \sin^{2} 15^{\circ} + \cos^{2} 15^{\circ}$.
Using the trigonometric identity $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$= 1$.

(C) We are given the expression: $\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ}$.
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we can rewrite the terms:
$\sin ^{2} 87^{\circ} = \sin ^{2}(90^{\circ} - 3^{\circ}) = \cos ^{2} 3^{\circ}$
$\sin ^{2} 89^{\circ} = \sin ^{2}(90^{\circ} - 1^{\circ}) = \cos ^{2} 1^{\circ}$
Substituting these into the original expression:
$= \sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ} + \cos ^{2} 1^{\circ}$
Grouping the terms with the same angle:
$= (\sin ^{2} 1^{\circ} + \cos ^{2} 1^{\circ}) + (\sin ^{2} 3^{\circ} + \cos ^{2} 3^{\circ})$
Using the identity $\sin ^{2} \theta + \cos ^{2} \theta = 1$:
$= 1 + 1 = 2$.

(D) Given equation: $3 \sin \theta = 4 \cos \theta$
Divide both sides by $\cos \theta$ (assuming $\cos \theta \neq 0$):
$\frac{3 \sin \theta}{\cos \theta} = 4$
Now,divide both sides by $3$:
$\frac{\sin \theta}{\cos \theta} = \frac{4}{3}$
Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get:
$\tan \theta = \frac{4}{3}$

(C) We know that the reciprocal of $\tan \theta$ is $\cot \theta$,so $\frac{1}{\tan \theta} = \cot \theta$.
Therefore,$\frac{1}{\tan^{2} \theta} = \cot^{2} \theta$.
Substituting this into the expression,we get $\cot^{2} \theta + 1$.
Using the trigonometric identity $1 + \cot^{2} \theta = \operatorname{cosec}^{2} \theta$,the final result is $\operatorname{cosec}^{2} \theta$.

(A) Given that $A$ and $B$ are acute angles.
Since $\tan A = 1,$ we know that $\tan 45^{\circ} = 1,$ therefore $A = 45^{\circ}.$
Since $\sin B = \frac{1}{\sqrt{2}},$ we know that $\sin 45^{\circ} = \frac{1}{\sqrt{2}},$ therefore $B = 45^{\circ}.$
Now,we need to find the value of $\cos (A + B).$
Substituting the values of $A$ and $B,$ we get $\cos (45^{\circ} + 45^{\circ}) = \cos 90^{\circ}.$
Since $\cos 90^{\circ} = 0,$ the final answer is $0$.