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Question 2:

We know
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$
Squaring on both sides, we get

$\therefore \frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=\frac{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=1$

Question 3:

We know
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$
Squaring on both sides, we get

$\therefore \frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=\frac{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{1-2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }=1$

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Question 4:

Ans

$\left(\mathrm{sec}A-\mathrm{cos}A\right)\left(\mathrm{cot}A+\mathrm{tan}A\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{\mathrm{cos}A}-\mathrm{cos}A\right)\left(\frac{\mathrm{cos}A}{\mathrm{sin}A}+\frac{\mathrm{sin}A}{\mathrm{cos}A}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-{\mathrm{cos}}^{2}A}{\mathrm{cos}A}\right)\left(\frac{{\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A}{\mathrm{sin}A\mathrm{cos}A}\right)$

Question 5:

$\left(\mathrm{sec}A-\mathrm{cos}A\right)\left(\mathrm{cot}A+\mathrm{tan}A\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{\mathrm{cos}A}-\mathrm{cos}A\right)\left(\frac{\mathrm{cos}A}{\mathrm{sin}A}+\frac{\mathrm{sin}A}{\mathrm{cos}A}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1-{\mathrm{cos}}^{2}A}{\mathrm{cos}A}\right)\left(\frac{{\mathrm{cos}}^{2}A+{\mathrm{sin}}^{2}A}{\mathrm{sin}A\mathrm{cos}A}\right)$

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Question 9:

$\frac{\mathrm{cosec}A-\mathrm{sin}A}{\mathrm{cosec}A+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{\mathrm{sin}A}-\mathrm{sin}A}{\frac{1}{\mathrm{sin}A}+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1-{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}{\frac{1+{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}\phantom{\rule{0ex}{0ex}}=\frac{1-{\mathrm{sin}}^{2}A}{1+{\mathrm{sin}}^{2}A}$
$=\frac{\frac{1-{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}{\frac{1+{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}$           (Dividing numerator and denominator by cos2A)

Question 10:

$\frac{\mathrm{cosec}A-\mathrm{sin}A}{\mathrm{cosec}A+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1}{\mathrm{sin}A}-\mathrm{sin}A}{\frac{1}{\mathrm{sin}A}+\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}=\frac{\frac{1-{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}{\frac{1+{\mathrm{sin}}^{2}A}{\mathrm{sin}A}}\phantom{\rule{0ex}{0ex}}=\frac{1-{\mathrm{sin}}^{2}A}{1+{\mathrm{sin}}^{2}A}$
$=\frac{\frac{1-{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}{\frac{1+{\mathrm{sin}}^{2}A}{{\mathrm{cos}}^{2}A}}$           (Dividing numerator and denominator by cos2A)

${\mathrm{sin}}^{2}\theta \mathrm{tan}\theta +{\mathrm{cos}}^{2}\theta \mathrm{cot}\theta +2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{3}\theta }{\mathrm{cos}\theta }+\frac{{\mathrm{cos}}^{3}\theta }{\mathrm{sin}\theta }+2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta +2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)}^{2}}{\mathrm{sin}\theta \mathrm{cos}\theta }$

$=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\theta +\mathrm{cot}\theta$

Question 11:

${\mathrm{sin}}^{2}\theta \mathrm{tan}\theta +{\mathrm{cos}}^{2}\theta \mathrm{cot}\theta +2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{3}\theta }{\mathrm{cos}\theta }+\frac{{\mathrm{cos}}^{3}\theta }{\mathrm{sin}\theta }+2\mathrm{sin}\theta \mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta +2{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta }{\mathrm{sin}\theta \mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{{\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)}^{2}}{\mathrm{sin}\theta \mathrm{cos}\theta }$

$=\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=\mathrm{tan}\theta +\mathrm{cot}\theta$

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

Question 12:

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

$\left(1+\mathrm{cot}A+\mathrm{tan}A\right)\left(\mathrm{sin}A-\mathrm{cos}A\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\mathrm{cot}A\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{tan}A\mathrm{cos}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\frac{\mathrm{cos}A}{\mathrm{sin}A}×\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\frac{\mathrm{sin}A}{\mathrm{cos}A}×\mathrm{cos}A$
$=\mathrm{sin}A+\mathrm{cos}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A\mathrm{tan}A-\mathrm{cot}A\mathrm{cos}A$

Question 13:

$\left(1+\mathrm{cot}A+\mathrm{tan}A\right)\left(\mathrm{sin}A-\mathrm{cos}A\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\mathrm{cot}A\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{tan}A\mathrm{cos}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A+\frac{\mathrm{cos}A}{\mathrm{sin}A}×\mathrm{sin}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\frac{\mathrm{sin}A}{\mathrm{cos}A}×\mathrm{cos}A$
$=\mathrm{sin}A+\mathrm{cos}A+\mathrm{tan}A\mathrm{sin}A-\mathrm{cos}A-\mathrm{cot}A\mathrm{cos}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}=\mathrm{sin}A\mathrm{tan}A-\mathrm{cot}A\mathrm{cos}A$

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ANS

ANS

Hence, LHS= RHS

Question 23:

Hence, LHS= RHS

Hence, L.H.S. = R.H.S.

Question 24:

Hence, L.H.S. = R.H.S.

ANS

ANS

ANS

ANS

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Ans

Hence, LHS = RHS

Hence, LHS = RHS

Hence, LHS = RHS

Hence, LHS = RHS

Also,

We know

Also,

We know

Question 9:

Given:

x = secA + sinA       .....(1)

y = secA – sinA       .....(2)

Adding (1) and (2), we get

$x+y=\mathrm{sec}A+\mathrm{sin}A+\mathrm{sec}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}A=x+y\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A=\frac{x+y}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}A}=\frac{2}{x+y}$

Subtracting (2) from (1), we get

We know

Question 10:

Given:

x = secA + sinA       .....(1)

y = secA – sinA       .....(2)

Adding (1) and (2), we get

$x+y=\mathrm{sec}A+\mathrm{sin}A+\mathrm{sec}A-\mathrm{sin}A\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}A=x+y\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A=\frac{x+y}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}A}=\frac{2}{x+y}$

Subtracting (2) from (1), we get

We know

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

Question 11:

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

$\mathrm{cos}\theta -\mathrm{sin}\theta =\sqrt{2}\mathrm{sin}\theta$
Squaring on both sides, we get

Question 12:

$\mathrm{cos}\theta -\mathrm{sin}\theta =\sqrt{2}\mathrm{sin}\theta$
Squaring on both sides, we get

$\mathrm{sec}\theta -\mathrm{tan}\theta =\sqrt{2}\mathrm{tan}\theta$
Squaring on both sides, we get

Question 13:

$\mathrm{sec}\theta -\mathrm{tan}\theta =\sqrt{2}\mathrm{tan}\theta$
Squaring on both sides, we get

Given:
We know

Adding (1) and (2), we get
$\mathrm{sec}\theta +\mathrm{tan}\theta +\mathrm{sec}\theta -\mathrm{tan}\theta =p+\frac{1}{p}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}\theta =\frac{{p}^{2}+1}{p}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}\theta =\frac{{p}^{2}+1}{2p}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}\theta }=\frac{2p}{{p}^{2}+1}$

Subtracting (2) from (1), we get

Now,

Question 14:

Given:
We know

Adding (1) and (2), we get
$\mathrm{sec}\theta +\mathrm{tan}\theta +\mathrm{sec}\theta -\mathrm{tan}\theta =p+\frac{1}{p}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}\theta =\frac{{p}^{2}+1}{p}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}\theta =\frac{{p}^{2}+1}{2p}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{sec}\theta }=\frac{2p}{{p}^{2}+1}$

Subtracting (2) from (1), we get

Now,

Given:
We know

Adding (1) and (2), we get
$\mathrm{cosec}A+\mathrm{cot}A+\mathrm{cosec}A-\mathrm{cot}A=m+\frac{1}{m}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cosec}A=\frac{{m}^{2}+1}{m}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cosec}A=\frac{{m}^{2}+1}{2m}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cosec}A}=\frac{2m}{{m}^{2}+1}$

Subtracting (2) from (1), we get

Now,

Question 15:

Given:
We know

Adding (1) and (2), we get
$\mathrm{cosec}A+\mathrm{cot}A+\mathrm{cosec}A-\mathrm{cot}A=m+\frac{1}{m}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cosec}A=\frac{{m}^{2}+1}{m}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cosec}A=\frac{{m}^{2}+1}{2m}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cosec}A}=\frac{2m}{{m}^{2}+1}$

Subtracting (2) from (1), we get

Now,

Given:
We know

Adding (1) and (2), we get

Subtracting (1) from (2), we get

Dividing (3) by (4), we get
$\frac{\frac{1+{x}^{2}}{2x}}{\frac{1-{x}^{2}}{2x}}=\frac{\mathrm{sec}A}{\mathrm{tan}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{\frac{1}{\mathrm{cos}A}}{\frac{\mathrm{sin}A}{\mathrm{cos}A}}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{1}{\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\mathrm{cosec}A$

Question 1:

Given:
We know

Adding (1) and (2), we get

Subtracting (1) from (2), we get

Dividing (3) by (4), we get
$\frac{\frac{1+{x}^{2}}{2x}}{\frac{1-{x}^{2}}{2x}}=\frac{\mathrm{sec}A}{\mathrm{tan}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{\frac{1}{\mathrm{cos}A}}{\frac{\mathrm{sin}A}{\mathrm{cos}A}}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{1}{\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\mathrm{cosec}A$

(b) 1

Question 2:

(b) 1

Hence, the correct answer is option (c).

Question 3:

Hence, the correct answer is option (c).

(b) cos A

Question 4:

(b) cos A

$\left(1+\mathrm{tan}\theta +\mathrm{sec}\theta \right)\left(1+\mathrm{cot}\theta -\mathrm{cosec}\theta \right)\phantom{\rule{0ex}{0ex}}=1+\mathrm{cot}\theta -\mathrm{cosec}\theta +\mathrm{tan}\theta +\mathrm{tan}\theta \mathrm{cot}\theta -\mathrm{tan}\theta \mathrm{cosec}\theta +\mathrm{sec}\theta +\mathrm{sec}\theta \mathrm{cot}\theta -\mathrm{sec}\theta \mathrm{cosec}\theta \phantom{\rule{0ex}{0ex}}=1+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\mathrm{tan}\theta ×\frac{1}{\mathrm{tan}\theta }-\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }×\frac{1}{\mathrm{sin}\theta }+\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{cos}\theta }×\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }$
$=1+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+1-\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=2+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=2+\frac{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{\mathrm{cos}\theta \mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }$

Hence, the correct answer is option (d).

Question 5:

$\left(1+\mathrm{tan}\theta +\mathrm{sec}\theta \right)\left(1+\mathrm{cot}\theta -\mathrm{cosec}\theta \right)\phantom{\rule{0ex}{0ex}}=1+\mathrm{cot}\theta -\mathrm{cosec}\theta +\mathrm{tan}\theta +\mathrm{tan}\theta \mathrm{cot}\theta -\mathrm{tan}\theta \mathrm{cosec}\theta +\mathrm{sec}\theta +\mathrm{sec}\theta \mathrm{cot}\theta -\mathrm{sec}\theta \mathrm{cosec}\theta \phantom{\rule{0ex}{0ex}}=1+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+\mathrm{tan}\theta ×\frac{1}{\mathrm{tan}\theta }-\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }×\frac{1}{\mathrm{sin}\theta }+\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{cos}\theta }×\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }$
$=1+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }-\frac{1}{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }+1-\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{cos}\theta }+\frac{1}{\mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=2+\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }+\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }\phantom{\rule{0ex}{0ex}}=2+\frac{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta }{\mathrm{cos}\theta \mathrm{sin}\theta }-\frac{1}{\mathrm{cos}\theta \mathrm{sin}\theta }$

Hence, the correct answer is option (d).

$\mathrm{sin}\theta -\mathrm{cos}\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\mathrm{tan}45°$
$⇒\theta =45°\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta \phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{4}45°+{\mathrm{cos}}^{4}45°\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}\right)}^{4}+{\left(\frac{1}{\sqrt{2}}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}+\frac{1}{4}$
$=\frac{2}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Hence, the correct answer is option (b).

Question 6:

$\mathrm{sin}\theta -\mathrm{cos}\theta =0\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\mathrm{tan}45°$
$⇒\theta =45°\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{sin}}^{4}\theta +{\mathrm{cos}}^{4}\theta \phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{4}45°+{\mathrm{cos}}^{4}45°\phantom{\rule{0ex}{0ex}}={\left(\frac{1}{\sqrt{2}}\right)}^{4}+{\left(\frac{1}{\sqrt{2}}\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}+\frac{1}{4}$
$=\frac{2}{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}$

Hence, the correct answer is option (b).

$\mathrm{cos}9\alpha =\mathrm{sin}\alpha \phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}9\alpha =\mathrm{cos}\left(90°-\alpha \right)\phantom{\rule{0ex}{0ex}}⇒9\alpha =90°-\alpha \phantom{\rule{0ex}{0ex}}⇒10\alpha =90°$
$⇒5\alpha =\frac{90°}{2}=45°\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}5\alpha =\mathrm{tan}45°=1$

Thus, the value of tan5α is 1.

Hence, the correct answer is option (c).

Question 7:

$\mathrm{cos}9\alpha =\mathrm{sin}\alpha \phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}9\alpha =\mathrm{cos}\left(90°-\alpha \right)\phantom{\rule{0ex}{0ex}}⇒9\alpha =90°-\alpha \phantom{\rule{0ex}{0ex}}⇒10\alpha =90°$
$⇒5\alpha =\frac{90°}{2}=45°\phantom{\rule{0ex}{0ex}}\therefore \mathrm{tan}5\alpha =\mathrm{tan}45°=1$

Thus, the value of tan5α is 1.

Hence, the correct answer is option (c).

Hence, the correct answer is option (d).

Question 8:

Hence, the correct answer is option (d).

(b) (sec A − tan A)

Question 9:

(b) (sec A − tan A)

$\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}}=\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}×\frac{1+\mathrm{cos}\left(A\right)}{1+\mathrm{cos}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{1-{\mathrm{cos}}^{2}\left(A\right)}}\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{{\mathrm{sin}}^{2}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}=\frac{1}{\mathrm{sin}\left(A\right)}+\frac{\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left(A\right)+\mathrm{cot}\left(A\right).$
Hence, the correct answer is option B.

Question 10:

$\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}}=\sqrt{\frac{1+\mathrm{cos}\left(A\right)}{1-\mathrm{cos}\left(A\right)}×\frac{1+\mathrm{cos}\left(A\right)}{1+\mathrm{cos}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{1-{\mathrm{cos}}^{2}\left(A\right)}}\phantom{\rule{0ex}{0ex}}=\frac{1+\mathrm{cos}\left(A\right)}{\sqrt{{\mathrm{sin}}^{2}\left(A\right)}}=\frac{1+\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}=\frac{1}{\mathrm{sin}\left(A\right)}+\frac{\mathrm{cos}\left(A\right)}{\mathrm{sin}\left(A\right)}\phantom{\rule{0ex}{0ex}}=\mathrm{cosec}\left(A\right)+\mathrm{cot}\left(A\right).$
Hence, the correct answer is option B.

(b) $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}$

${\left(\text{cosec}\theta -\text{cot}\theta \right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}{\left(\frac{1}{\text{sin}\theta }-\frac{\text{cos}\theta }{\text{sin}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}{\left(\frac{1-\text{cos}\theta }{\text{sin}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{{\text{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{\left(1-{\text{cos}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{\left(1+\text{cos}\theta \right)\left(1-\text{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left(1-\text{cos}\theta \right)}{\left(1+\text{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}$

Question 11:

(b) $\frac{1-\mathrm{cos\theta }}{1+\mathrm{cos\theta }}$

${\left(\text{cosec}\theta -\text{cot}\theta \right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}{\left(\frac{1}{\text{sin}\theta }-\frac{\text{cos}\theta }{\text{sin}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}{\left(\frac{1-\text{cos}\theta }{\text{sin}\theta }\right)}^{2}\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{{\text{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{\left(1-{\text{cos}}^{2}\theta \right)}\phantom{\rule{0ex}{0ex}}\text{=}\frac{{\left(1-\text{cos}\theta \right)}^{2}}{\left(1+\text{cos}\theta \right)\left(1-\text{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left(1-\text{cos}\theta \right)}{\left(1+\text{cos}\theta \right)}\phantom{\rule{0ex}{0ex}}$

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25       [âˆµ tan θ = ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23

Question 12:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25       [âˆµ tan θ = ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23

(c) $\frac{b+a}{b-a}$

Question 13:

(c) $\frac{b+a}{b-a}$

(d) $\frac{3}{4}$

=

Question 14:

(d) $\frac{3}{4}$

=

Now,

Hence, the correct answer is option (a).

Question 15:

Now,

Hence, the correct answer is option (a).

Hence, the correct answer is option (c).

Question 16:

Hence, the correct answer is option (c).

In âˆ†ABC,

∠A + ∠B + ∠C = 180º                (Angle sum property of triangle)

⇒ ∠A + ∠B + 90º = 180º            (∠C = 90º)

⇒ ∠A + ∠B = 180º − 90º = 90º

∴ cos(A + B) = cos90º = 0

Thus, the value of cos(A + B) is 0.

Hence, the correct answer is option (a).

Question 17:

In âˆ†ABC,

∠A + ∠B + ∠C = 180º                (Angle sum property of triangle)

⇒ ∠A + ∠B + 90º = 180º            (∠C = 90º)

⇒ ∠A + ∠B = 180º − 90º = 90º

∴ cos(A + B) = cos90º = 0

Thus, the value of cos(A + B) is 0.

Hence, the correct answer is option (a).

(a) 1

Question 18:

(a) 1

(d) 9

We have .

Dividing the numerator and denominator of the given expression by sin θ, we get:

=

= = 9              [âˆµ 3 cot θ = 4]

Question 19:

(d) 9

We have .

Dividing the numerator and denominator of the given expression by sin θ, we get:

=

= = 9              [âˆµ 3 cot θ = 4]

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 2
= 2
=
= $\frac{1}{2}$                  [By using the identity: ]

Question 20:

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 2
= 2
=
= $\frac{1}{2}$                  [By using the identity: ]

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 3
= 3
=
= $\frac{1}{3}$       [By using the identity: ]

Question 21:

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = and .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3$\left({x}^{2}-\frac{1}{{x}^{2}}\right)$ = 3
= 3
=
= $\frac{1}{3}$       [By using the identity: ]

(d) 2

Let us first draw a right $∆$ABC right angled at B and $\angle \mathrm{A}=\theta$.
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{BC}{AB}$
So, $\frac{BC}{AB}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$k and AB = k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = ($\sqrt{3}$k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ =

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