Textbook - Introduction to Trigonometry Questions in English

(N/A) We start with the Left Hand Side $(LHS)$:
$LHS = \frac{\cot A - \cos A}{\cot A + \cos A}$
Substitute $\cot A = \frac{\cos A}{\sin A}$:
$LHS = \frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A}$
Factor out $\cos A$ from the numerator and the denominator:
$LHS = \frac{\cos A \left( \frac{1}{\sin A} - 1 \right)}{\cos A \left( \frac{1}{\sin A} + 1 \right)}$
Cancel $\cos A$ from both:
$LHS = \frac{\frac{1}{\sin A} - 1}{\frac{1}{\sin A} + 1}$
Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we substitute it:
$LHS = \frac{\operatorname{cosec} A - 1}{\operatorname{cosec} A + 1} = RHS$
Hence,the identity is proved.

(A) To prove the identity,we divide the numerator and the denominator of the $LHS$ by $\cos \theta$ to express it in terms of $\tan \theta$ and $\sec \theta.$
$LHS = \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}$
Rearranging the terms,we get:
$LHS = \frac{(\tan \theta + \sec \theta) - 1}{(\tan \theta - \sec \theta) + 1}$
Using the identity $\sec^2 \theta - \tan^2 \theta = 1,$ we substitute $1$ in the numerator:
$LHS = \frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{(\tan \theta - \sec \theta) + 1}$
Factorizing the numerator using $a^2 - b^2 = (a - b)(a + b)$:
$LHS = \frac{(\tan \theta + \sec \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{(\tan \theta - \sec \theta) + 1}$
Taking $(\tan \theta + \sec \theta)$ as a common factor:
$LHS = \frac{(\tan \theta + \sec \theta) [1 - (\sec \theta - \tan \theta)]}{(\tan \theta - \sec \theta) + 1}$
$LHS = \frac{(\tan \theta + \sec \theta) [1 - \sec \theta + \tan \theta]}{(\tan \theta - \sec \theta + 1)}$
Canceling the common term $(\tan \theta - \sec \theta + 1)$:
$LHS = \tan \theta + \sec \theta$
To get the $RHS,$ multiply and divide by $(\sec \theta - \tan \theta)$:
$LHS = \frac{(\sec \theta + \tan \theta)(\sec \theta - \tan \theta)}{\sec \theta - \tan \theta} = \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = \frac{1}{\sec \theta - \tan \theta} = RHS.$

(N/A) We know that,$\operatorname{cosec}^{2} A = 1 + \cot^{2} A$.
Since $\sin A = \frac{1}{\operatorname{cosec} A}$,we have $\sin^{2} A = \frac{1}{\operatorname{cosec}^{2} A} = \frac{1}{1 + \cot^{2} A}$.
Taking the square root,$\sin A = \frac{1}{\sqrt{1 + \cot^{2} A}}$.
Next,we know that $\tan A = \frac{1}{\cot A}$.
Finally,using the identity $\sec^{2} A = 1 + \tan^{2} A$,we substitute $\tan A = \frac{1}{\cot A}$:
$\sec^{2} A = 1 + \left(\frac{1}{\cot A}\right)^{2} = 1 + \frac{1}{\cot^{2} A} = \frac{\cot^{2} A + 1}{\cot^{2} A}$.
Taking the square root,$\sec A = \frac{\sqrt{1 + \cot^{2} A}}{\cot A}$.

We know that,$\cos A = \frac{1}{\sec A}$.
Using the identity $\sin^2 A + \cos^2 A = 1$,we have $\sin^2 A = 1 - \cos^2 A = 1 - \frac{1}{\sec^2 A} = \frac{\sec^2 A - 1}{\sec^2 A}$.
Thus,$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$.
Using the identity $\tan^2 A + 1 = \sec^2 A$,we have $\tan^2 A = \sec^2 A - 1$.
Thus,$\tan A = \sqrt{\sec^2 A - 1}$.
Since $\cot A = \frac{1}{\tan A}$,we have $\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$.
Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we have $\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$.

(C) Given expression: $\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
Using the complementary angle identities $\sin(90^{\circ} - \theta) = \cos \theta$ and $\cos(90^{\circ} - \theta) = \sin \theta$:
Numerator: $\sin ^{2} 63^{\circ} + \sin ^{2} 27^{\circ} = [\sin(90^{\circ} - 27^{\circ})]^{2} + \sin ^{2} 27^{\circ} = \cos ^{2} 27^{\circ} + \sin ^{2} 27^{\circ} = 1$
Denominator: $\cos ^{2} 17^{\circ} + \cos ^{2} 73^{\circ} = [\cos(90^{\circ} - 73^{\circ})]^{2} + \cos ^{2} 73^{\circ} = \sin ^{2} 73^{\circ} + \cos ^{2} 73^{\circ} = 1$
Therefore,the value is $\frac{1}{1} = 1$.

(D) The given expression is $\sin 25^{\circ} \cos 65^{\circ} + \cos 25^{\circ} \sin 65^{\circ}$.
We know that $\cos(90^{\circ} - \theta) = \sin \theta$ and $\sin(90^{\circ} - \theta) = \cos \theta$.
Substituting $\cos 65^{\circ} = \cos(90^{\circ} - 25^{\circ}) = \sin 25^{\circ}$ and $\sin 65^{\circ} = \sin(90^{\circ} - 25^{\circ}) = \cos 25^{\circ}$:
$= (\sin 25^{\circ})(\sin 25^{\circ}) + (\cos 25^{\circ})(\cos 25^{\circ})$
$= \sin^{2} 25^{\circ} + \cos^{2} 25^{\circ}$
Using the identity $\sin^{2} A + \cos^{2} A = 1$,we get:
$= 1$

(A) Given expression: $9 \sec^{2} A - 9 \tan^{2} A$
Factor out the common term $9$:
$= 9(\sec^{2} A - \tan^{2} A)$
Using the trigonometric identity $\sec^{2} A - \tan^{2} A = 1$:
$= 9(1)$
$= 9$

(B) Given expression: $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$
Convert the trigonometric ratios into $\sin \theta$ and $\cos \theta$:
$= \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)$
Simplify the terms inside the brackets:
$= \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)$
Apply the algebraic identity $(a+b)(a-b) = a^2 - b^2$,where $a = (\sin \theta + \cos \theta)$ and $b = 1$:
$= \frac{(\sin \theta + \cos \theta)^2 - (1)^2}{\sin \theta \cos \theta}$
Expand the numerator using $(a+b)^2 = a^2 + b^2 + 2ab$:
$= \frac{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta}$
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$= \frac{1 + 2 \sin \theta \cos \theta - 1}{\sin \theta \cos \theta}$
$= \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta} = 2$

(C) Given expression: $(\sec A + \tan A)(1 - \sin A)$
Step $1$: Convert $\sec A$ and $\tan A$ into terms of $\sin A$ and $\cos A$:
$(\frac{1}{\cos A} + \frac{\sin A}{\cos A})(1 - \sin A)$
Step $2$: Combine the fractions:
$(\frac{1 + \sin A}{\cos A})(1 - \sin A)$
Step $3$: Multiply the numerators:
$\frac{(1 + \sin A)(1 - \sin A)}{\cos A}$
Step $4$: Use the identity $(a + b)(a - b) = a^2 - b^2$:
$\frac{1 - \sin^2 A}{\cos A}$
Step $5$: Use the identity $1 - \sin^2 A = \cos^2 A$:
$\frac{\cos^2 A}{\cos A} = \cos A$

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

(A) To prove: $(\operatorname{cosec} \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
Consider the $L.H.S. = (\operatorname{cosec} \theta - \cot \theta)^2$
Using the trigonometric identities $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta}$,we get:
$L.H.S. = \left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2$
$= \left( \frac{1 - \cos \theta}{\sin \theta} \right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we get:
$= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$
Since $a^2 - b^2 = (a - b)(a + b)$,we can write $1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$:
$= \frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)}$
$= \frac{1 - \cos \theta}{1 + \cos \theta} = R.H.S.$
Hence,$L.H.S. = R.H.S.$

(N/A) To prove: $\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A$
$L.H.S. = \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}$
Taking the common denominator:
$= \frac{\cos^2 A + (1+\sin A)^2}{(1+\sin A)(\cos A)}$
Expanding the numerator:
$= \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{(1+\sin A)(\cos A)}$
Using the identity $\sin^2 A + \cos^2 A = 1$:
$= \frac{(\sin^2 A + \cos^2 A) + 1 + 2\sin A}{(1+\sin A)(\cos A)}$
$= \frac{1 + 1 + 2\sin A}{(1+\sin A)(\cos A)}$
$= \frac{2 + 2\sin A}{(1+\sin A)(\cos A)}$
Factoring out $2$:
$= \frac{2(1+\sin A)}{(1+\sin A)(\cos A)}$
$= \frac{2}{\cos A} = 2 \sec A$
$= R.H.S.$

(A) $L.H.S. = \frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$= \frac{1}{(\sin \theta-\cos \theta)} \left[ \frac{\sin^2 \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta} \right]$
$= \left( \frac{1}{\sin \theta-\cos \theta} \right) \left[ \frac{\sin^3 \theta-\cos^3 \theta}{\sin \theta \cos \theta} \right]$
$= \left( \frac{1}{\sin \theta-\cos \theta} \right) \left[ \frac{(\sin \theta-\cos \theta)(\sin^2 \theta+\cos^2 \theta+\sin \theta \cos \theta)}{\sin \theta \cos \theta} \right]$
$= \frac{1+\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \operatorname{cosec} \theta \sec \theta + 1 = R.H.S.$

(A) To prove: $\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$
Step $1$: Simplify the Left Hand Side $(L.H.S.)$.
$L.H.S. = \frac{1+\sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$
Step $2$: Simplify the fraction.
$= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \cos A + 1$
Step $3$: Simplify the Right Hand Side $(R.H.S.)$ to match the $L.H.S.$ or multiply the $L.H.S.$ by $\frac{1-\cos A}{1-\cos A}$.
$= (1 + \cos A) \times \frac{1-\cos A}{1-\cos A}$
Step $4$: Use the identity $\sin^2 A + \cos^2 A = 1$,so $1 - \cos^2 A = \sin^2 A$.
$= \frac{1 - \cos^2 A}{1 - \cos A} = \frac{\sin^2 A}{1 - \cos A}$
Thus,$L.H.S. = R.H.S.$

(N/A) To prove: $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$
$L.H.S. = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$
Divide the numerator and the denominator by $\sin A$:
$= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$
$= \frac{\cot A - 1 + \operatorname{cosec} A}{\cot A + 1 - \operatorname{cosec} A}$
Using the identity $1 = \operatorname{cosec}^{2} A - \cot^{2} A$ in the numerator:
$= \frac{\cot A + \operatorname{cosec} A - (\operatorname{cosec}^{2} A - \cot^{2} A)}{\cot A - \operatorname{cosec} A + 1}$
$= \frac{(\cot A + \operatorname{cosec} A) - (\operatorname{cosec} A - \cot A)(\operatorname{cosec} A + \cot A)}{\cot A - \operatorname{cosec} A + 1}$
Factor out $(\cot A + \operatorname{cosec} A)$:
$= \frac{(\cot A + \operatorname{cosec} A) [1 - (\operatorname{cosec} A - \cot A)]}{\cot A - \operatorname{cosec} A + 1}$
$= \frac{(\cot A + \operatorname{cosec} A) (1 - \operatorname{cosec} A + \cot A)}{\cot A - \operatorname{cosec} A + 1}$
Since $(1 - \operatorname{cosec} A + \cot A) = (\cot A - \operatorname{cosec} A + 1)$,they cancel out:
$= \cot A + \operatorname{cosec} A = R.H.S.$

(N/A) To prove the identity $\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$:
$L.H.S. = \sqrt{\frac{1+\sin A}{1-\sin A}}$
Multiply the numerator and denominator inside the square root by $(1 + \sin A)$:
$= \sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$
Using the identity $(1 - \sin^2 A) = \cos^2 A$:
$= \sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}} = \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}$
Taking the square root:
$= \frac{1+\sin A}{\cos A}$
$= \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A$
$= R.H.S.$

(A) Given: $L.H.S. = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$
Step $1$: Factor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator.
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$
Step $2$: Use the identity $\sin^2 \theta + \cos^2 \theta = 1$,which implies $\cos^2 \theta = 1 - \sin^2 \theta$.
Substitute this into the denominator:
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta [2(1 - \sin^2 \theta) - 1]}$
Step $3$: Simplify the expression inside the bracket in the denominator.
$2(1 - \sin^2 \theta) - 1 = 2 - 2 \sin^2 \theta - 1 = 1 - 2 \sin^2 \theta$
Step $4$: Substitute back into the expression.
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (1 - 2 \sin^2 \theta)}$
Step $5$: Cancel the common term $(1 - 2 \sin^2 \theta)$.
$L.H.S. = \frac{\sin \theta}{\cos \theta} = \tan \theta = R.H.S.$
Hence,the identity is proved.

(N/A) To prove: $(\sin A + \operatorname{cosec} A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
$L.H.S. = (\sin A + \operatorname{cosec} A)^2 + (\cos A + \sec A)^2$
Using the identity $(a + b)^2 = a^2 + b^2 + 2ab$:
$= (\sin^2 A + \operatorname{cosec}^2 A + 2 \sin A \operatorname{cosec} A) + (\cos^2 A + \sec^2 A + 2 \cos A \sec A)$
$= (\sin^2 A + \cos^2 A) + \operatorname{cosec}^2 A + \sec^2 A + 2 \sin A \left(\frac{1}{\sin A}\right) + 2 \cos A \left(\frac{1}{\cos A}\right)$
Since $\sin^2 A + \cos^2 A = 1$,$\operatorname{cosec}^2 A = 1 + \cot^2 A$,and $\sec^2 A = 1 + \tan^2 A$:
$= 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 2(1) + 2(1)$
$= 1 + 1 + \cot^2 A + 1 + \tan^2 A + 2 + 2$
$= 7 + \tan^2 A + \cot^2 A$
$= R.H.S.$

(A) To prove: $(\operatorname{cosec} A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}$
$L.H.S. = (\operatorname{cosec} A - \sin A)(\sec A - \cos A)$
$= (\frac{1}{\sin A} - \sin A)(\frac{1}{\cos A} - \cos A)$
$= (\frac{1 - \sin^2 A}{\sin A})(\frac{1 - \cos^2 A}{\cos A})$
$= (\frac{\cos^2 A}{\sin A})(\frac{\sin^2 A}{\cos A})$
$= \sin A \cos A$
$R.H.S. = \frac{1}{\tan A + \cot A}$
$= \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}}$
$= \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}}$
$= \frac{\sin A \cos A}{\sin^2 A + \cos^2 A}$
$= \frac{\sin A \cos A}{1} = \sin A \cos A$
Since $L.H.S. = R.H.S.$,the identity is proved.

(A) First,consider the expression $\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$.
Using the identities $1+\tan ^{2} A = \sec ^{2} A$ and $1+\cot ^{2} A = \operatorname{cosec}^{2} A$,we get:
$\frac{\sec ^{2} A}{\operatorname{cosec}^{2} A} = \frac{1/\cos ^{2} A}{1/\sin ^{2} A} = \frac{\sin ^{2} A}{\cos ^{2} A} = \tan ^{2} A$.
Next,consider the expression $\left(\frac{1-\tan A}{1-\cot A}\right)^{2}$.
Substitute $\cot A = \frac{1}{\tan A}$:
$\left(\frac{1-\tan A}{1-\frac{1}{\tan A}}\right)^{2} = \left(\frac{1-\tan A}{\frac{\tan A - 1}{\tan A}}\right)^{2} = \left(\frac{(1-\tan A) \cdot \tan A}{-(1-\tan A)}\right)^{2} = (-\tan A)^{2} = \tan ^{2} A$.
Since both parts equal $\tan ^{2} A$,the identity is proved.