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#### Page No 546:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that sin $\theta$ = $\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$= $\frac{AB}{AC}$ = .

So, if AB = $\sqrt{3}k$, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 $-$ AB2 = (2k)2 $-$ ($\sqrt{3}k$)2
⇒ BC2 = 4k2 $-$ 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos $\theta$  = $\frac{BC}{AC}$ =
tan $\theta$  =

∴ cot $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Page No 546:

Let us first draw a right $∆$ABC, right angled at B and .
Now, we know that cos $\theta$ = = = .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 $-$ BC2 = (25k)2 $-$ (7k)2.
⇒ AB2 = 625k2 $-$ 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin $\theta$ = =
tan $\theta$ =
∴ cot $\theta$ = , cosec $\theta$ = and sec $\theta$  =

#### Page No 546:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$ = $\frac{\mathrm{Perpendicular}}{\mathrm{Base}}$ = $\frac{AB}{BC}$ = $\frac{15}{8}$.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
sin $\theta$  = $\frac{AB}{AC}$ =
cos $\theta$  =

∴ cot $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Page No 546:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cot $\theta$$\frac{\mathrm{base}}{\mathrm{Perpendicular}}$ = $\frac{BC}{AB}$ = 2.

So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = $\sqrt{5}$k
Now, finding the other T-ratios using their definitions, we get:
sin $\theta$  = $\frac{AB}{AC}$ =
cos $\theta$  =

∴ tan $\theta$  = , cosec $\theta$ = and sec $\theta$  =

#### Page No 546:

Let us first draw a right $∆$ABC, right angled at B and $\angle C=\theta$.
Now, we know that cosec $\theta$ = $\frac{\mathrm{Hypotenuse}}{\mathrm{Perpendicular}}$ = $\frac{AC}{AB}$= $\frac{\sqrt{10}}{1}$.

So, if AC = ($\sqrt{10}$)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 $-$ AB2 = 10k2 $-$ k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan $\theta$  = $\frac{AB}{BC}$ =

cos $\theta$  =

∴ , cot $\theta$  = and sec $\theta$  =

#### Page No 546:

We have $\mathrm{sin}\theta =\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}$,

As,

${\mathrm{cos}}^{2}\theta =1-{\mathrm{sin}}^{2}\theta \phantom{\rule{0ex}{0ex}}=1-{\left(\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{1}-\frac{{\left({a}^{2}-{b}^{2}\right)}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left({a}^{2}+{b}^{2}\right)}^{2}-{\left({a}^{2}-{b}^{2}\right)}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left[\left({a}^{2}+{b}^{2}\right)-\left({a}^{2}-{b}^{2}\right)\right]\left[\left({a}^{2}+{b}^{2}\right)+\left({a}^{2}-{b}^{2}\right)\right]}{{\left({a}^{2}+{b}^{2}\right)}^{2}}$
$=\frac{\left[{a}^{2}+{b}^{2}-{a}^{2}+{b}^{2}\right]\left[{a}^{2}+{b}^{2}+{a}^{2}-{b}^{2}\right]}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left[2{b}^{2}\right]\left[2{a}^{2}\right]}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{2}\theta =\frac{4{a}^{2}{b}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\sqrt{\frac{4{a}^{2}{b}^{2}}{{\left({a}^{2}+{b}^{2}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{2ab}{\left({a}^{2}+{b}^{2}\right)}$

Also,

$\mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}\right)}{\left(\frac{2ab}{{a}^{2}+{b}^{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}-{b}^{2}}{2ab}$

Now,

$\mathrm{cosec}\theta =\frac{1}{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{{a}^{2}-{b}^{2}}{{a}^{2}+{b}^{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}+{b}^{2}}{{a}^{2}-{b}^{2}}$

Also,

$\mathrm{sec}\theta =\frac{1}{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{2ab}{{a}^{2}+{b}^{2}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{2}+{b}^{2}}{2ab}$

And,

$\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\frac{{a}^{2}-{b}^{2}}{2ab}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2ab}{{a}^{2}-{b}^{2}}$

#### Page No 546:

$\mathrm{LHS}=\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{\mathrm{cos}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{sin}\theta }{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{sin}\theta }{\sqrt{1-{\mathrm{sin}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\frac{a}{b}\right)}{\sqrt{1-{\left(\frac{a}{b}\right)}^{2}}}$
$=\frac{\left(\frac{1}{1}+\frac{a}{b}\right)}{\sqrt{\frac{1}{1}-\frac{{a}^{2}}{{b}^{2}}}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{b+a}{b}\right)}{\sqrt{\frac{{b}^{2}-{a}^{2}}{{b}^{2}}}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{b+a}{b}\right)}{\left(\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(b+a\right)}{\sqrt{\left(b+a\right)\left(b-a\right)}}$
$=\frac{\left(b+a\right)}{\sqrt{\left(b+a\right)}\sqrt{\left(b-a\right)}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{\left(b+a\right)}}{\sqrt{\left(b-a\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{b+a}{b-a}}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Page No 547:

It is given that tan .

LHS =
Dividing the numerator and denominator by cos $\theta$, we get:

(∵ tan )
Now, substituting the value of tan $\theta$ in the above expression, we get:

i.e., LHS = RHS

Hence proved.

#### Page No 547:

$\mathrm{LHS}=\left(3\mathrm{cos}\alpha -4{\mathrm{cos}}^{3}\alpha \right)\phantom{\rule{0ex}{0ex}}=\mathrm{cos}\alpha \left(3-4{\mathrm{cos}}^{2}\alpha \right)\phantom{\rule{0ex}{0ex}}=\sqrt{1-{\mathrm{sin}}^{2}\alpha }\left[3-4\left(1-{\mathrm{sin}}^{2}\alpha \right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{1-{\left(\frac{1}{2}\right)}^{2}}\left[3-4\left(1-{\left(\frac{1}{2}\right)}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{1}-\frac{1}{4}}\left[3-4\left(\frac{1}{1}-\frac{1}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{4}}\left[3-4\left(\frac{3}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{4}}\left[3-3\right]\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{4}}\left[0\right]\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Page No 547:

It is given that cot .

LHS  =
Dividing the above expression by sin $\theta$, we get:
[∵ cot ]
Now, substituting the values of cot $\theta$ in the above expression, we get:

i.e., LHS = RHS

Hence proved.

#### Page No 547:

It is given that sec $\theta$ = $\frac{17}{8}$.

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
We know that cos $\theta$ =

So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 $-$ BC2 = (17k)2 $-$ (8k)2
⇒ AB2 = 289k2 $-$ 64k2 = 225k2
⇒ AB = 15k.

Now, tan $\theta$  = and sin $\theta$ =

The given expression is .

Substituting the values in the above expression, we get:

∴ LHS = RHS
Hence proved.

#### Page No 547:

Let us consider a right $△$ABC right angled at B and $\angle C=\theta$.
Now, we know that tan $\theta$$\frac{AB}{BC}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒  AC = 29k
Now, sin $\theta$ = and cos $\theta$ =

Substituting these values in the given expression, we get:

∴ LHS = RHS

Hence proved.

#### Page No 547:

Let us consider a right $△$ABC, right angled at B and $\angle C=\theta$.
Now it is given that tan $\theta$$\frac{AB}{BC}$$\frac{1}{\sqrt{7}}$.

So, if AB = k, then BC = $\sqrt{7}$k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + ($\sqrt{7}$k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 2$\sqrt{2}$k
Now, finding out the values of the other trigonometric ratios, we have:
sin $\theta$  =
cos $\theta$  =
∴ cosec $\theta$  = and sec $\theta$   =
Substituting the values of cosec $\theta$  and sec $\theta$  in the given expression, we get:

i.e., LHS = RHS

Hence proved.

#### Page No 547:

$\mathrm{LHS}=\sqrt{\frac{{\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta }{{\mathrm{sec}}^{2}\theta -1}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{{\mathrm{tan}}^{2}\theta }}\phantom{\rule{0ex}{0ex}}=\sqrt{{\mathrm{cot}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{cot}\theta \phantom{\rule{0ex}{0ex}}=\sqrt{{\mathrm{cosec}}^{2}\theta -1}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{1}{\mathrm{sin}\theta }\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{1}{\left(\frac{3}{4}\right)}\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{4}{3}\right)}^{2}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{16}{9}-1}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{16-9}{9}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{7}{9}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}}{3}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Page No 547:

(i)
​
$=\sqrt{\frac{\left(\frac{1}{4}\right)}{\left(\frac{7}{4}\right)}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{7}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

(ii)

#### Page No 547:

$\mathrm{LHS}={\left(\frac{2}{x+y}\right)}^{2}+{\left(\frac{x-y}{2}\right)}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\frac{2}{\left(\mathrm{cosecA}+\mathrm{cosA}\right)+\left(\mathrm{cosecA}-\mathrm{cosA}\right)}\right]}^{2}+{\left[\frac{\left(\mathrm{cosecA}+\mathrm{cosA}\right)-\left(\mathrm{cosecA}-\mathrm{cosA}\right)}{2}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\frac{2}{\mathrm{cosecA}+\mathrm{cosA}+\mathrm{cosecA}-\mathrm{cosA}}\right]}^{2}+{\left[\frac{\mathrm{cosecA}+\mathrm{cosA}-\mathrm{cosecA}+\mathrm{cosA}}{2}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\frac{2}{2\mathrm{cosecA}}\right]}^{2}+{\left[\frac{2\mathrm{cosA}}{2}\right]}^{2}-1$
$={\left[\frac{1}{\mathrm{cosecA}}\right]}^{2}+{\left[\mathrm{cosA}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\left[\mathrm{sinA}\right]}^{2}+{\left[\mathrm{cosA}\right]}^{2}-1\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{2}\mathrm{A}+{\mathrm{cos}}^{2}\mathrm{A}-1\phantom{\rule{0ex}{0ex}}=1-1\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}=\mathrm{RHS}$

#### Page No 548:

In ,

Using Pythagoras theorem, we get

$\mathrm{PQ}=\sqrt{{\mathrm{PR}}^{2}-{\mathrm{QR}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(x+2\right)}^{2}-{x}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{x}^{2}+4x+4-{x}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4\left(x+1\right)}\phantom{\rule{0ex}{0ex}}=2\sqrt{x+1}$

Now,

#### Page No 548:

In $∆$ABC, $\angle$C = 90$°$
sinA = $\frac{\mathrm{BC}}{\mathrm{AB}}$ and
sinB = $\frac{\mathrm{AC}}{\mathrm{AB}}$

As, sinA = sinB
$⇒$$\frac{\mathrm{BC}}{\mathrm{AB}}$ = $\frac{\mathrm{AC}}{\mathrm{AB}}$
$⇒$BC = AC
So, $\angle$A = $\angle$B             (Angles opposite to equal sides are equal)

In ,

#### Page No 555:

Hence, the correct option is (c).

#### Page No 555:

Hence, the correct option is (a).

#### Page No 555:

Hence, the correct option is (d).

#### Page No 555:

Hence, the correct option is (b).

#### Page No 555:

Hence, the correct option is (c).

#### Page No 555:

Hence, the correct option is (a).

#### Page No 556:

Hence, the correct option is (c).