Question 13 Marks
Prove the given identities, where the angles involved are acute angles for which the expressions are defined.$(\sin A+ cosec\ A)^2+(\cos A+\sec A)^2=7+\tan ^2 A+\cot ^2 A$
AnswerTo prove: $(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^2+(\cos \mathrm{A}+\sec \mathrm{A})^2=7+\tan ^2 \mathrm{~A}+\cot ^2 \mathrm{~A}$
taking L.H.S
Using the formula $(a+b)^2=a^2+b^2+2 a b$ to get,
$=\left(\sin ^2 A+\operatorname{cosec}^2 A+2 \sin A \operatorname{cosec} A\right)+\left(\cos ^2 A+\sec ^2 A+2 \cos A \sec A\right)$
Since $\sin \theta=\frac{1}{\operatorname{cosec} \theta}$ and $\cos \theta=\frac{1}{\sec \theta}$
$ =\left(\sin ^2 A+\csc ^2 A+2 \sin A \frac{\operatorname{cosec} \theta}{\sin A}\right)+\left(\cos ^2 A+\sec ^2 A+2 \cos A \frac{1}{\cos A}\right)$
$ =\sin ^2 \mathrm{~A}+\operatorname{cosec}^2 \mathrm{~A}+2+\cos ^2 \mathrm{~A}+\sec ^2 \mathrm{~A}+2 $
$ =\left(\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right)+\operatorname{cosec}^2 \mathrm{~A}+\sec ^2 \mathrm{~A}+2+2$
Using the identities $\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}=1, \sec ^2 \mathrm{~A}=1+\tan ^2 \mathrm{~A}$ and $\operatorname{cosec}^2 \mathrm{~A}=1+\cot ^2 \mathrm{~A}$ to get
$=1+1+\tan ^2 \mathrm{~A}+1+\cot ^2 \mathrm{~A}+2+2$
$ =1+2+2+2+\tan ^2 \mathrm{~A}+\cot ^2 \mathrm{~A} $
$ =7+\tan ^2 \mathrm{~A}+\cot ^2 \mathrm{~A}$
= R.H.S.
Hence proved
View full question & answer→Question 23 Marks
Prove the given identity, where the angles involved are acute angles for which the expressions are defined.$\frac{\cos A-\sin A+1}{\cos A+\sin A-1} = cosec\ A + \cot A$, using the identity $\operatorname{cosec}^2 \mathrm{~A}=1+\cot ^2 \mathrm{~A}$
AnswerTaking L.H.S
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$
Dividing Numerator and Denominator by $\sin A$
$=\frac{\frac{\cos A-\sin A+1}{\sin A}}{\frac{\cos A+\sin A-1}{\sin A}}$
$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}-\frac{1}{\sin A}}$
Using the formula $\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}$$=\frac{\cot A-1+\ cosec A}{\cot A+1-\ cosec A}$
Using the identity $\operatorname{cosec}^2 \mathrm{~A}=1+\cot ^2 \mathrm{~A}$
$=\frac{\cot A-\left(\ cosec ^{2} A-\cot ^{2} A\right)+\ cosec A}{\cot A+1-\ cosec A}$
$=\frac{(\cot A+\ cosec A)-\left(\ cosec ^{2} A-\cot ^{2} A\right)}{\cot A+1-\ cosec A}$
$=\frac{(\cot A+\ cosec A)(1-\ cosec A+\cot A)}{\cot A+1-\ cosec A}$
$= \cot A + cosec\ A$
$=$ R.H.S
View full question & answer→Question 33 Marks
Prove the given identity, where the angles involved are acute angles for which the expressions are defined.$ \frac{{\tan A }}{{1 - \cot A }} + \frac{{\cot A }}{{1 - \tan A }} = 1 + \sec A \ cosec\ A$
$[$Hint: Write the expression in terms of $\sin \theta$ and $\cos \theta]$
AnswerLHS- $\frac{{\tan A}}{{1 - \cot A }} + \frac{{\cot A }}{{1 - \tan A }}$
$=\frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} + \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}}$
$=\frac{{\tan A}}{{\frac{{\tan A - 1}}{{\tan A}}}} + \frac{1}{{\tan A(1 - \tan A)}}$
$=\frac{{{{\tan }^2}A}}{{\tan A - 1}} + \frac{1}{{\tan A(1 - \tan A)}}$
$=\frac{{{{\tan }^3}A - 1}}{{\tan A(\tan A - 1)}}$
$=\frac{{(\tan A - 1)({{\tan }^2}A + \tan A + 1)}}{{\tan A(\tan A - 1)}} $
$ \left[a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=\frac{{{{\tan }^2}A + \tan A + 1}}{{\tan A}}$
$=\tan A + 1 + \cot A$
$=\frac{{\sin A}}{{\cos A}} + \frac{{\cos A}}{{\sin A}} + 1$
$=\frac{{{{\sin }^2}A + {{\cos }^2}A}}{{\sin A\cos A}} + 1$
$=\frac{1}{{\sin A\cos A}} + 1$
$= \sec A\ cosec\ A+ 1$
$=$ R.H.S
View full question & answer→Question 43 Marks
Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A.$
Answer$\sin A$ can be expressed in terms of $\sec A$ as:
$\sin A=\sqrt{\sin ^{2} A}$
$\sin A=\sqrt{( 1-\cos ^{2} A)}$
$\sin A=\sqrt{1-\frac{1}{\sec ^{2} A}}$
$\sin A=\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}$
$\sin A=\frac{1}{\sec A} \sqrt{\sec ^{2} A-1}$
Now,
$\cos A$ can be expressed in terms of $\sec A$ as:
$\cos A=\frac{1}{\sec A}$
$\tan A$ can be expressed in the form of $\sec A$ as:
As,$1+\tan ^2 \mathrm{~A}=\sec ^2 \mathrm{~A}$
$\left.\Rightarrow \tan A=\pm \sqrt{( \sec ^{2} A-1}\right)$
since $A$ is an acute angle, and $\tan A$ is positive when $A$ is acute, So, $\tan A = \sqrt{( \sec ^{2} A-1)}$
Now $\ cosec A$ can be expressed in the form of $\sec A$ as:
$\ cosec \ A=\frac{1}{\sin A}$
$\ cosec\ A=\frac{1}{\frac{\sec A}{\sqrt{1-\sec ^{2} A}}} $
$\ cosec\ A=\frac{\sqrt{1-\sec ^{2} A}}{\sec A} $ Now, $\cot A$ can be expressed in terms of $\sec A$ as:
$\cot A =\frac{1}{\tan A}$
as,$1+\tan ^2 \mathrm{~A}=\sec ^2 \mathrm{~A}$
$\ \cot A=\frac{1}{\sqrt{\sec ^{2} A-1}}$
View full question & answer→Question 53 Marks
Express the trigonometric ratios $\sin \mathrm{A}, \sec \mathrm{A}$ and $\tan \mathrm{A}$ in terms of $\cot \mathrm{A}$.
AnswerWe have to express the trigonometric ratios $\sin A, \sec A$ and $\tan A$ in terms of $\cot A.$
For $\sin A,$ By using identity $cosec ^ { 2 } A - \cot ^ { 2 } A = 1$
$\Rightarrow cosec ^ { 2 } A = 1 + \cot ^ { 2 } A$
$\Rightarrow \frac { 1 } { \sin ^ { 2 } A } = 1 + \cot ^ { 2 } A$
$\Rightarrow \sin ^ { 2 } A = \frac { 1 } { 1 + \cot ^ { 2 } A }$
$ \Rightarrow \quad \sin A = \frac { 1 } { \sqrt { 1 + \cot ^ { 2 } A } }$
For $\sec A,$ By using identity $ \sec ^ { 2 } A - \tan ^ { 2 } A = 1$
$ \Rightarrow \sec ^ { 2 } A = 1 + \tan ^ { 2 } A$
$ \Rightarrow \sec ^ { 2 } A = 1 + \frac { 1 } { \cot ^ { 2 } A } = \frac { \cot ^ { 2 } A + 1 } { \cot ^ { 2 } A }$
$ \Rightarrow \sec ^ { 2 } A = \frac { 1 + \cot ^ { 2 } A } { \cot ^ { 2 } A }$
$ \Rightarrow \sec A = \frac { \sqrt { 1 + \cot ^ { 2 } A } } { \cot A }$
For $\tan A, \tan A = \frac { 1 } { \cot A }$
View full question & answer→Question 63 Marks
In $\triangle A B C$, right angled at $B,$ if $\tan A = \frac { 1 } { \sqrt { 3 } }$. Find the value of $\cos A \cos C - \sin A \sin C$
Answer
we have,
$\tan A = \frac { 1 } { \sqrt { 3 } }$
$= \tan 30^\circ$
$\therefore A = 30^\circ$
In $\triangle ABC$, we have
$\angle A+\angle B+\angle C=180^\circ$
$\Rightarrow30^\circ+90^\circ+\angle C=180^\circ$
$\Rightarrow120^\circ+\angle C=180^\circ$
$\Rightarrow\angle C=180^\circ-120^\circ=60^\circ$
So,
$\cos A.\cos C - \sin A.\sin C$
$=\cos30^\circ.\cos60^\circ-\sin30^\circ.\sin60^\circ$
$=\frac{\sqrt3}2\cdot\frac12-\frac12\cdot\frac{\sqrt3}2=0$ View full question & answer→Question 73 Marks
In $\triangle A B C$, right angled at $B,$ if $\tan A = \frac { 1 } { \sqrt { 3 } }$. Find the value of $\sin A \cos C + \cos A \sin C.$
Answer
we have,
$\tan A = \frac { 1 } { \sqrt { 3 } }$
$= \tan 30^\circ$
$\therefore A = 30^\circ$
In $\triangle ABC,$ we have
$\angle A+\angle B+\angle C=180^\circ$
$\Rightarrow30^\circ+90^\circ+\angle C=180^\circ$
$\Rightarrow120^\circ+\angle C=180^\circ$
$\Rightarrow\angle C=180^\circ-120^\circ=60^\circ$
So,
$\sin A . \cos C + \cos A . \sin C$
$=\sin30^\circ.\cos60^\circ+\cos30^\circ.\sin60^\circ$
$=\frac12\cdot\frac12+\frac{\sqrt3}2\cdot\frac{\sqrt3}2=1$ View full question & answer→Question 83 Marks
If $3 \cot A = 4,$ check whether $\frac { 1 - \tan ^ { 2 } A } { 1 + \tan ^ { 2 } A } = \cos ^ { 2 } A - \sin ^ { 2 } A$ or not.
AnswerGive that $3 \cot A = 4$
Or $\cot A = \frac{4}{3}$
Consider a right angle triangle $\Delta ABC$ right angled at point $B.$
$\cot A = \frac{{Side\;adjacent\;to\;\angle A}}{{Side\;opposite\;to\;\angle A}}$
$\frac{{AB}}{{BC}} = \frac{4}{3}$
If $AB$ is $4K, BC$ will be $3K$. where $K$ is a positive integer
Now in $\Delta ABC$
$(A C)^2=(A B)^2+(B C)^2$
$=(4 K)^2+(3 K)^2 $
$=16 K^2+9 K^2 $
$=25 K^2$
$AC = 5K$
$\cos A = \frac{{Side\;adjacent\;to\;\angle A}}{{hypotenuse}} = \frac{{AB}}{{AC}}$
$ = \frac{{4K}}{{5K}} = \frac{4}{5}$
$\sin A = \frac{{Side\;opposite\;to\;\angle A}}{{hypotenuse}} = \frac{{BC}}{{AC}}$
$ = \frac{{3K}}{{5K}} = \frac{3}{5}$
$\tan A = \frac{{Side\;opposite\;to\;\angle A}}{{Side\;adjacent\;to\;angle\;A}} = \frac{{BC}}{{AB}}$
$ = \frac{{3K}}{{4K}} = \frac{3}{4}$
$\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}} = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}$
$ = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} = \frac{7}{{25}}$
${\cos ^2}A - {\sin ^2}A = {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}$
$ = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{7}{{25}}$
Hence $\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A$ View full question & answer→Question 93 Marks
If $\cot \theta = \frac { 7 } { 8 }$, evaluate: $\cot ^ { 2 } \theta$
AnswerLet us consider a right angled $\Delta ABC$ right angled at point $B.$
Let $\angle C = \theta$
Given,
$\cot \theta = \frac {7} {8} =\frac{{Side\;adjacent\;to\;\angle \theta }}{{Side\;opposite\;to\;\angle \theta }} = \frac{{\mathrm{BC}}}{{\mathrm{AB}}}$
If $BC$ is $7K$ then $AB$ will be $8K,$ where $K$ is a positive integer.
Now applying Pythagoras theorem in $\Delta ABC$
$A C^2=A B^2+B C^2$
Or, $A C^2=64 \mathrm{~K}^2+49 \mathrm{~K}^2$
Or, $A C^2=113 K^2$
$\therefore \mathrm{AC} = \sqrt {113} \mathrm{K}$
Now,
$\sin \theta = \frac{{Side\;opposite\;to\;\angle \theta }}{{hypotenuse}} = \frac{{\mathrm{AB}}}{{\mathrm{AC}}}$
$ = \frac{{8K}}{{\sqrt {113} K}} = \frac{8}{{\sqrt {113} }}$
And,
$\cos \theta = \frac{{Side\;adjacent\;to\;\angle \theta }}{{hypotenuse}} = \frac{{\mathrm{BC}}}{{\mathrm{AC}}}$
$ = \frac{{7K}}{{\sqrt {113} K}} = \frac{7}{{\sqrt {113} }}$
Now ${\cot ^2}\theta = {\left( {\frac{\cos \theta}{\sin \theta}} \right)^2} ={\left( {\frac{7}{{\sqrt {113} }}\times\frac{\sqrt {113}}{{ {8} }}} \right)^2}= {\left( {\frac{7}{{ {8} }}} \right)^2}=\frac{{49}}{{64}}$ View full question & answer→Question 103 Marks
Given $\sec \theta = \frac { 13 } { 12 }$, Calculate all other trigonometric ratios.
AnswerConsider a triangle $ABC$ in which $\angle A = \theta $ and $\angle B = {90^o}$
Let $AB = 12k$ and $AC = 13k$
Then, using Pythagoras theorem,
$BC=\sqrt { ( \mathrm { AC } ) ^ { 2 } - ( \mathrm { AB } ) ^ { 2 } } = \sqrt { ( 13 k ) ^ { 2 } - ( 12 k ) ^ { 2 } }$
$= \sqrt { 169 k ^ { 2 } - 144 k ^ { 2 } } = \sqrt { 25 k ^ { 2 } } = 5 k$
$\therefore $ $\sin \theta = \frac { B C } { A C } = \frac { 5 k } { 13 k } = \frac { 5 } { 13 }$
$\cos \theta = \frac { A B } { A C } = \frac { 12 k } { 13 k } = \frac { 12 } { 13 } \tan \theta = \frac { B C } { A B } = \frac { 5 k } { 12 k } = \frac { 5 } { 12 }$
$\cot \theta = \frac { A B } { B C } = \frac { 12 k } { 5 k } = \frac { 12 } { 5 } \cos e c \theta = \frac { A C } { B C } = \frac { 13 k } { 5 k } = \frac { 13 } { 5 }$ View full question & answer→Question 113 Marks
Given $15 \cot A = 8$ find $\sin A$ and $\sec A.$
AnswerLet us draw a right triangle $ABC.$
$15 \cot A = 8 ......$ Given
$\Rightarrow \cot A = \frac { 8 } { 15 } \Rightarrow \frac { A B } { B C } = \frac { 8 } { 15 }$
$\Rightarrow \frac { A B } { 8 } = \frac { B C } { 15 } = k ( 5 a y )$
where $k$ is a positive number
$\Rightarrow A B = 8 k$
$BC = 15k$
By using the Pythagoras theorem, we have
$A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }$
$\Rightarrow A C ^ { 2 } = ( 8 k ) ^ { 2 } + ( 15 k ) ^ { 2 } \Rightarrow A C ^ { 2 } = 64 k ^ { 2 } + 225 k ^ { 2 }$
$\Rightarrow A C = \sqrt { 289 k ^ { 2 } } \Rightarrow A C = 17 k$
Now, $\sin A = \frac { B C } { A C } = \frac { 15 k } { 17 k } = \frac { 15 } { 17 }$
and, $\sec A = \frac { A C } { A B } = \frac { 17 k } { 8 k } = \frac { 17 } { 8 }$ View full question & answer→Question 123 Marks
In $ \triangle P Q R$, right angled at $Q, PR + QR = 25\ cm$ and $PQ = 5\ cm$. Determine the values of $\sin P, \cos P$ and $\tan P.$
Answer
In $ \triangle P Q R$ by Pythagoras theorem
$\mathrm{PR}^2=\mathrm{PQ}^2+\mathrm{QR}^2 $
$\Rightarrow(25-Q R)^2=5^2+Q R^2[\because P R+Q R=25 \mathrm{~cm} \Rightarrow P R=25-\mathrm{QR}] $
$625-50 \mathrm{QR}+\mathrm{QR}^2=25+\mathrm{QR}^2$
$ \Rightarrow 600 - 50 Q R = 0$
$ \Rightarrow Q R = \frac { 600 } { 50 } = 12 \;\mathrm { cm }$
Now, $PR + QR = 25 cm$
$ \Rightarrow PR = 25 - Q R = 25 - 12 = 13 cm$
Hence,$\sin P = \frac { Q R } { P R } = \frac { 12 } { 13 } , \cos P = \frac { P Q } { P R } = \frac { 5 } { 13 } \text { and, } \tan P = \frac { Q R } { P Q } = \frac { 12 } { 5 }$ View full question & answer→Question 133 Marks
In $ \triangle ABC,$ right angled at $B, AB = 24\ cm, BC = 7\ cm.$ Determine:
- $\sin A \cos A$
- $\sin C \cos C$
AnswerLet us draw a right angled triangle $ABC,$ right angled at $B.$
Using Pythagoras theorem,
Given, $AB = 24\ cm$ and $BC = 7\ cm$
Using Pythagoras theorem,
$AC^2=AB^2+BC^2$
$AC^2$ = $24^2+7^2=576+49=625$
$\therefore AC = 25\ cm$
- $\sin \mathrm { A } = \frac { \mathrm { P } } { \mathrm { H } } = \frac { \mathrm { BC } } { \mathrm { AC } } = \frac { 7 } { 25 },$
$\cos A = \frac { B } { H } = \frac { A B } { A C } = \frac { 24 } { 25 }$
$\Rightarrow \sin A.\cos A$ = $\frac{7}{25} \times \frac{24}{25} = \frac{168}{625}$ ii. $\sin \mathrm { C } = \frac { \mathrm { P } } { \mathrm { H } } = \frac { \mathrm { AB } } { \mathrm { AC } } = \frac { 24 } { 25 },$
$\cos C = \frac { B } { H } = \frac { B C } { A C } = \frac { 7 } { 25 }$
$\Rightarrow \sin C.\cos C$ = $\frac{24}{25} \times \frac{7}{25} = \frac{168}{625}$ View full question & answer→Question 143 Marks
In $\triangle ABC,$ right-angled at $B, AB = 5\ cm$ and $\angle ACB = 30^\circ .$ Determine the lengths os sides $BC$ and $AC.$
AnswerGiven $AB = 5\ cm $
$\angle ACB = 30^\circ$
According to diagram,
$\tan C = \frac{side \ opposite \ to \ angle \ C}{side \ adjacent \ to \ angle \ C}$
$\tan 30^\circ= \frac{AB}{BC}$
$\frac{1}{\sqrt{3}}$ = $\frac{5}{BC}$
$BC = 5\sqrt{3} cm$
$\sin C = \frac{side \ of \ angle \ C}{hypotenuse}$
$\sin 30^\circ= \frac{AB}{AC}$
$\frac{1}{2}$ = $\frac{5}{AC}$
$AC = 10\ cm.$ View full question & answer→Question 153 Marks
In $\triangle O P Q$ right angled at $P, OP = 7\ cm, OQ - PQ = 1\ cm.$ Determine the values of $\sin Q$ and $\cos Q.$
Answer
In, $\triangle O P Q$ by Pythagoras theorem
$O Q^2=O P^2+P Q^2$
$\Rightarrow ( P Q + 1 ) ^ { 2 } = O P ^ { 2 } + P Q ^ { 2 } \ [ \because O Q - P Q = 1 \Rightarrow O Q = 1 + P Q ]$
$\Rightarrow P Q ^ { 2 } + 2 P Q + 1 = 7 ^ { 2 } + P Q ^ { 2 }$
$\Rightarrow 2 P Q + 1 = 49$
$\Rightarrow 2 P Q = 48$
$\Rightarrow P Q = 24\;\mathrm { cm }$
$\therefore O Q - P Q = 1 \mathrm { cm }$
$\Rightarrow OQ - 24 = 1$
$\Rightarrow OQ = 25\ cm$
Now, $\sin Q = \frac { O P } { O Q } = \frac { 7 } { 25 }$
and, $\cos Q = \frac { P Q } { O Q } = \frac { 24 } { 25 }$ View full question & answer→Question 163 Marks
In a right triangle $ABC,$ right-angled at $B,$ if $\tan A = 1,$ then verify that $2 \sin A \cos A = 1.$
Answer
In $ \triangle A B C$,
$\tan A = 1$
$\Rightarrow \quad \frac { B C } { A C } = 1$
$\Rightarrow BC = x$ and $AC = x$
Using Pythagoras theorem,
$\Rightarrow A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }$
$\Rightarrow \quad A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }$
$\Rightarrow \quad A B = \sqrt { 2 } x$
$\therefore \quad \sin A = \frac { B C } { A B } = \frac { x } { \sqrt { 2 } x } = \frac { 1 } { \sqrt { 2 } } \text { and } \cos A = \frac { A C } { \sqrt { 2 } x } = \frac { x } { \sqrt { 2 } x } = \frac { 1 } { \sqrt { 2 } } 2 \sin A \cos A = 2 \times \frac { 1 } { \sqrt { 2 } } \times \frac { 1 } { \sqrt { 2 } } = 1$ View full question & answer→Question 173 Marks
Consider $ \triangle A C B$ right angled at $C$ in which $AB = 29$ units, $BC = 21$ units and $ \angle A B C = \theta$. Determine the values of
- $ \cos ^2 \theta+\sin ^2 \theta $
- $ \cos ^2 \theta-\sin ^2 \theta$
AnswerIn, $ \Delta A C B$ we have
$ A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }$
$ \Rightarrow \quad A C = \sqrt { A B ^ { 2 } - B C ^ { 2 } } = \sqrt { 29 ^ { 2 } - 21 ^ { 2 } } = \sqrt { ( 29 + 21 ) ( 29 - 21 ) } = \sqrt { 400 } = 20$ units
$ \therefore \quad \sin \theta = \frac { A C } { A B } = \frac { 20 } { 29 }$and $ \cos \theta = \frac { B C } { A B } = \frac { 21 } { 29 }$
- Using the values of $ \sin \theta$ and,$ \cos \theta$ we get $ \cos ^ { 2 } \theta + \sin ^ { 2 } \theta = \left( \frac { 21 } { 29 } \right) ^ { 2 } + \left( \frac { 20 } { 29 } \right) ^ { 2 }$
$ = \frac { 441 + 400 } { 841 } = 1$
- Using the values of $ \sin \theta$ and, $ \cos \theta$ we obtain
$\cos^2\theta-\sin^2\theta=\left(\frac{21}{29}\right)^2-\left(\frac{20}{29}\right)^2=\frac{21^2-20^2}{29^2}=\frac{(21+20)(21-20)}{841}=\frac{41}{841}$
View full question & answer→Question 183 Marks
If $ \angle B $ and $\angle Q$ are acute angles such that $\sin B = \sin Q,$ then prove that $ \angle B = \angle Q$.
AnswerConsider two right triangles $ABC$ and $PQR$ in which $ \angle B$ and $\angle Q$ are the right angles.
We have,
In $\triangle ABC$
$\sin B=\frac{AC}{AB}$
and, In $\triangle PQR$
$\sin Q=\frac{PR}{PQ}$
$ \because \quad \sin B = \sin Q$
$ \Rightarrow \quad \frac { A C } { A B } = \frac { P R } { P Q }$
$ \Rightarrow \quad \frac { A C } { P R } = \frac { A B } { P Q } = k($say$) ...... (i)$
$ \Rightarrow AC = kPR$ and $AB = kPQ .....(ii)$
Using Pythagoras theorem in triangles $ABC$ and $PQR,$ we obtain
$A B^2=A C^2+B C^2$ and $\mathrm{PQ}^2=\mathrm{PR}^2+\mathrm{QR}^2$
$ \Rightarrow \quad B C = \sqrt { A B ^ { 2 } - A C ^ { 2 } } \text { and } Q R = \sqrt { P Q ^ { 2 } - P R ^ { 2 } }$
$ \Rightarrow \quad \frac { B C } { Q R } = \frac { \sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \frac { \sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } [ $using $(ii) ]$
$ \Rightarrow \quad \frac { B C } { Q R } = \frac { k \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k...(iii)$
From $(i)$ and $(iii),$ we get
$ \frac { A C } { P R } = \frac { A B } { P Q } = \frac { B C } { Q R }$
$ \Rightarrow \quad \Delta A C B - \Delta P R Q [$By $S.A.S$ similarity$]$
$ \therefore \quad \angle B = \angle Q$
Hence proved. View full question & answer→Question 193 Marks
Prove that $\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}$, using identity $\sec^2\theta=1+ \tan^2\theta$.
AnswerWe have to prove that,$\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}$ using identity $\sec^2\theta=1+\tan^2\theta$
LHS = $\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} $
$ = \frac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}$ [ dividing the numerator and denominator by $\cos{\theta}$.] $ = \frac{{(\tan \theta + \sec \theta)-1 }}{{(\tan \theta - \sec \theta )+1}}$
$=\frac{\{{(\tan\theta+\sec\theta)-1\}}(\tan\theta-\sec\theta)}{\{{(\tan\theta-\sec\theta)+1\}}(\tan\theta-\sec\theta)}$ [ Multiplying and dividing by
$(\tan{\theta}-\sec{\theta})$]
$=\frac{{(\tan^2\theta-\sec^2\theta)-}(\tan\theta-\sec\theta)}{\{{(\tan\theta-\sec\theta)+1\}}(\tan\theta-\sec\theta)}$ [$\because (a-b)(a+b)=a^2-b^2$]
$ = \frac{{-1-\tan \theta + \sec \theta }}{{(\tan \theta - \sec \theta+1)(\tan{\theta}-\sec{\theta}) }}$
[$\because \tan^2\theta-\sec^2\theta=-1$]
$=\frac{-(\tan\theta-\sec\theta+1)}{(\tan\theta-\sec\theta+1)(\tan\theta-\sec\theta)}$$=\frac{-1}{\tan{\theta}-\sec{\theta}}$
$ = \frac{1}{{\sec \theta - \tan \theta }}$=RHS Hence Proved.
View full question & answer→Question 203 Marks
Given $\tan A = \frac{4}{3}$, find all other trigonometric ratios of the angle $A.$
AnswerLet us first draw a right $\triangle ABC$
Now, we have given that, tan $A=\frac{B C}{A B}=\frac{4}{3}$
Therefore, if $BC = 4k,$ then $AB = 3k,$ where $k$ is any positive integer.
Now, by using the Pythagoras Theorem, we have
$A C^2=A B^2+B C^2=(4 k)^2+(3 k)^2=25 k$
So, $AC = 5k$
Now, we can write all the trigonometric ratios using their definitions
$\sin A=\frac{B C}{A C}=\frac{4 k}{5 k}=\frac{4}{5}$
$\cos A=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}$
Therefore,$\cot A = \frac{1}{\tan A}=\frac{3}{4},cosec\ A=\frac{1}{\sin A}=\frac{5}{4} $ and $\sec A=\frac{1}{\cos A}=\frac{5}{3}$
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