# Introduction To Trigonometry Exercise 8.4

Q.1     Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Sol.       Consider a $\Delta ABC$, in which $\angle B = 90^\circ$.
For $\angle A$, we have
Base = AB, Perp = BC
and Hyp = AC
Therefore, $\cot A = {{Base} \over {Perp}} = {{AB} \over {BC}}$ $\Rightarrow {{AB} \over {BC}} = \cot A = {{\cot A} \over 1}$
Let AB = k cotA and BC = k.
By Pythagoras Theorem,
$AC = \sqrt {A{B^2} + B{C^2}} = \sqrt {{k^2}{{\cot }^2}A + {k^2}}$
$= k\sqrt {1 + {{\cot }^2}A}$
Therefore, $\sin A = {{Perp} \over {Base}} = {{BC} \over {AC}} = {k \over {k\sqrt {1 + {{\cot }^2}A} }}$
$= {1 \over {\sqrt {1 + {{\cot }^2}A} }}$
$\sec A = {{Hyp} \over {Base}} = {{AC} \over {AB}} = {{k\sqrt {1 + {{\cot }^2}A} } \over {k\cot A}}$
$= \sqrt {{{1 + {{\cot }^2}A} \over {\cot A}}}$

and, $\tan A = {{Perp} \over {Base}} = {{BC} \over {AB}} = {k \over {k\cot A}} = {1 \over {\cot A}}$

Q.2     Write the other trigonometric ratios of A in terms of secA.
Sol.        Consider a $\Delta ABC$, in which $\angle B = 90^\circ$.
For $\angle A,$ we have
Base = AB, Perp = BC
and Hyp = AC
Therefore, $\sec A = {{Hyp} \over {Base}} = {{AC} \over {AB}}$
$\Rightarrow {{AC} \over {AB}} = \sec A = {{\sec A} \over 1}$ $\Rightarrow {{AB} \over {AC}} = {1 \over {\sec A}} = {{{1 \over {\sec A}}} \over 1}$ (Note this step)
Let $AB = k\left( {{1 \over {\sec A}}} \right),AC = k(1)$
By phythagoras Theorem
$BC = \sqrt {A{C^2} + A{B^2}} = \sqrt {{k^2} - {k^2}\left( {{1 \over {{{\sec }^2}A}}} \right)}$
$= k\sqrt {{{{{\sec }^2}A - 1} \over {{{\sec }^2}A}}} = {{k\sqrt {{{\sec }^2}A - 1} } \over {{{\sec }^2}A}}$

Therefore, $\sin A = {{BC} \over {AC}} = {{k\sqrt {{{\sec }^2}A - 1} } \over {{{\sec A} \over k}}} = {{\sqrt {{{\sec }^2}A - 1} } \over {\sec A}}$
$\cos A = {{AB} \over {AC}}{{k\left( {{1 \over {\sec A}}} \right)} \over k} = {1 \over {\sec A}}$
$\tan A = {{BC} \over {AB}} = {{{{k\sqrt {{{\sec }^2}A - 1} } \over {\sec A}}} \over {k\left( {{1 \over {\sec A}}} \right)}} = \sqrt {{{\sec }^2}A - 1}$
$\cot A = {1 \over {\tan A}} = {1 \over {\sqrt {{{\sec }^2}A - 1} }}$
$\cos ecA = {1 \over {\sin A}} = {{\sec A} \over {\sqrt {{{\sec }^2}A - 1} }}$

Q.3     Evaluate :
(i) ${{{{\sin }^2}63^\circ + {{\sin }^2}27^\circ } \over {{{\cos }^2}17^\circ + {{\cos }^2}73^\circ }}$
(ii) sin 25° cos 65° + cos25° sin65° $[Since,sin\left( {90^\circ - \theta } \right) = \cos \theta ]$
Sol.       (i) Here, sin63° = sin(90° – 27°) = cos27°
and cos17° = cos(90° – 73°) = sin 73° $[Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]$
Therefore, ${{{{\sin }^2}63^\circ + {{\sin }^2}27^\circ} \over {{{\cos }^2}17^\circ + {{\cos }^2}73^\circ}} = {{{{\cos }^2}27^\circ + {{\sin }^2}27^\circ }\over {{{\sin }^2}73^\circ + {{\cos }^2}73^\circ}}$
$= {1 \over 1} = 1$
$[Since,{\cos ^2}A + {\sin ^2}A = 1]$

(ii) sin25°cos65°+cos25°sin65°
= sin(90°– 65°). cos65° + cos(90°– 65°) sin65°
= cos65° cos65° + sin 65° sin65°
$[Since,\sin \left( {90^\circ - \theta } \right) = \cos \theta ]$
= ${\cos ^2}65^\circ + {\sin ^2}65^\circ = 1$ $[Since,\cos \left( {90^\circ - \theta } \right) = \sin \theta ]$

Q.4     Choose the correct option. Justify your choice :
(i) $9{\sec ^2}A - 9{\tan ^2}A =$
(A) 1                                     (B) 9
(C) 8                                      (D) 0

(ii) (1+  tan$\theta$ + sec$\theta$)(1 + cos $\theta$ – cosec $\theta$) =
(A) 0                                     (B) 1
(C) 2                                     (D) None of these

(iii) (secA + tanA)(1 – sinA) =
(A) secA                              (B) sinA
(C) cosecA                           (D) cosA

(iv) ${{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}} =$
(A) ${{{\sec }^2}A}$                              (B) –1
(C) ${{{\cot }^2}A}$                               (D) none of these
Sol.       (i) (B), because
$9{\sec ^2}A - 9{\tan ^2}A = 9\left( {{{\sec }^2}A - {{\tan }^2}A} \right)$
= 9 × 1 = 9 $\left[ {Since,1 + {{\tan }^2}A = {{\sec }^2}A} \right]$

(ii) (C), because
$\left( {1 + \tan \theta + \sec \theta } \right)\left( {1 + \cot \theta - \cos ec\theta } \right)$
$= \left( {1 + {{\sin \theta } \over {\cos \theta }} + {1 \over {\cos \theta }}} \right)\left( {1 + {{\cos \theta } \over {\sin \theta }} - {1 \over {\sin \theta }}} \right)$
$= \left( {{{\cos \theta + \sin \theta + 1} \over {\cos \theta }}} \right)\left( {{{\sin \theta + \cos \theta - 1} \over {\sin \theta }}} \right)$
= ${{{{\left( {\cos \theta + \sin \theta} \right)}^2} - 1} \over {\sin \theta \cos \theta }}$ $[Since,\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}]$
=  ${{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta} \right) - 2\cos \theta \sin \theta - 1} \over {\sin \theta \cos \theta }}$ $[Since,{\sin ^2}\theta + {\cos ^2}\theta = 1]$
= ${{1 + 2\cos \theta \sin \theta - 1} \over {\sin \theta \cos \theta }} = {{2\cos \theta \sin \theta } \over {\sin \theta \cos \theta }} = 2$

(iii) (D), because
(secA + tanA) (1 – sinA)
= $\left( {{1 \over {\cos A}} + {{\sin A} \over {\cos A}}} \right)\left( {1 - \sin A} \right)$
= $\left( {{{1 + \sin A} \over {\cos A}}} \right)\left( {1 - \sin A} \right)$ $[Since,\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^2}]$
$= {{1 - {{\sin }^2}A} \over {\cos A}}{{{{\cos }^2}A} \over {\cos A}} = \cos A[Since,{\sin ^2}A + {\cos ^2}A = 1]$

(iv) (D), because
${{1 + {{\tan }^2}A} \over {1 - {{\cot }^2}A}} = {{1 + {{\tan }^2}A} \over {1 + {1 \over {{{\tan }^2}A}}}} = {{1 + {{\tan }^2}A} \over {{{{{\tan }^2}A + 1} \over {{{\tan }^2}A}}}}$
$= \left( {1 + {{\tan }^2}A} \right) \times {{{{\tan }^2}A} \over {1 + {{\tan }^2}A}} = {\tan ^2}A$

Q.5     Prove the following identities, where the angles involved are acute angles for which the expressions are defined :
(i) ${\left( {\cos ec\theta - \cot \theta } \right)^2} = {{1 - \cos \theta } \over {1 + \cos \theta }}$
(ii) ${{\cos A} \over {1 + \sin A}} + {{1 + \sin A} \over {\cos A}} = 2\sec A$
(iii) ${{\tan \theta } \over {1 - \cot \theta }} + {{\cot \theta } \over {1 - \tan \theta }} = 1 + \sec \theta \cos ec\theta$
(iv) ${{1 + \sec A} \over {\sec A}} = {{{{\sin }^2}A} \over {1 - \cos A}}$
(v) ${{\cos A - \sin A+ 1} \over {\cos A + \sin A - 1}} = \cos ecA + \cot A,$ using the identity $\cos e{c^2}A = 1 + {\cot ^2}A$
(vi) $\sqrt {{{1 + \sin A} \over {1 - \sin A}} = } \sec A + \tan A$
(vii) ${{\sin \theta - 2{{\sin }^3}\theta } \over {2{{\cos }^3}\theta - \cos \theta }} = \tan \theta$
(viii) ${{\rm{(sinA + cosecA)}}^2} + {{\rm{(cosA + secA)}}^2} = 7 + {\tan ^2}A + {\cot ^2}A$
(ix) (cosecA – sinA)(secA – cosA) = ${1 \over {\tan A + \cot A}}$
(x) $\left( {{{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}}} \right) = {\left( {{{1 - \tan A} \over {1 - {{\cot }^2}A}}} \right)^2} = {\tan ^2}A$
Sol.       (i) We have,
L.H.S. = ${{\rm{(cosec}}\theta - \cot \theta {\rm{ )}}^2}$
$= {\left( {{1 \over {\sin \theta }} - {{\cos \theta } \over {\sin \theta }}} \right)^2} = {\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)^2}$
$= {{{{\left( {1 - \cos \theta } \right)}^2}} \over {{{\sin }^2}\theta }} = {{{{\left( {1 - \cos \theta } \right)}^2}} \over {1 - {{\cos }^2}\theta }}$ $[Since,{\sin ^2}\theta = 1 - {\cos ^2}\theta ]$
$= {{{{\left( {1 - \cos \theta } \right)}^2}} \over {\left( {1 - \cos \theta } \right)\left( {1 + \cos \theta } \right)}} = {{1 - \cos \theta } \over {1 + \cos \theta }}$
R.H.S. $[Since,{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]$

(ii) We have,
L.H.S. = ${{\cos A} \over {1 + \sin A}} + {{1 + \sin A} \over {\cos A}}$
= ${{{{\cos }^2}A + {{\left( {1 + \sin A} \right)}^2}} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${{{{\cos }^2}A + 1 + 2\sin A + {{\sin }^2}A} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${{\left( {{{\cos }^2}A + {{\sin }^2}A} \right) + 1 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${{1 + 1 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}}[Since {\sin ^2}A + {\cos ^2}A = 1]$
= ${{2 + 2\sin A} \over {\cos A\left( {1 + \sin A} \right)}} = {{2\left( {1 + \sin A} \right)} \over {\cos A\left( {1 + \sin A} \right)}}$
= ${2 \over {\cos A}} = 2\sec A = R.H.S.$

(iii) We have,
L.H.S. = ${{\tan \theta } \over {1 - \cot \theta }} + {{\cot \theta } \over {1 - \tan \theta }}$
= ${{\tan \theta } \over {1 - {1 \over {\tan \theta }}}} + {{{1 \over {\tan \theta }}} \over {1 - \tan \theta }}$
$= {{\tan \theta } \over {{{\tan \theta - 1} \over {\tan \theta }}}} + {1 \over {\tan \theta \left( {1 - \tan \theta } \right)}}$

$= {{{{\tan }^2}\theta } \over {\tan \theta - 1}} + {1 \over {\tan \theta \left( {1 - \tan \theta } \right)}}$
$= {{{{\tan }^2}\theta } \over {\tan \theta - 1}} - {1 \over {\tan \theta \left( {\tan \theta - 1} \right)}}$
= ${{{{\tan }^3}\theta - 1} \over {\tan \theta \left( {\tan \theta - 1} \right)}}$
= ${{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + \tan \theta + 1} \right)} \over {{{\tan }}\theta \left( {\tan \theta - 1} \right)}}$
$[Since,{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)]$
= ${{{{\tan }^2}\theta + \tan \theta + 1} \over {{{\tan }^2}\theta }}$
= ${{{{\tan }^2}\theta } \over {\tan \theta }} + {{\tan \theta } \over {\tan \theta }} + {1 \over {\tan \theta }}$
= $\tan \theta + 1 + \cot \theta = 1 + \tan \theta + \cot \theta$
= $1 + {{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}$
= $1 + {{{{\sin }^2}\theta + {{\cos }^2}\theta } \over {\cos \theta }}$
= $1 + {1 \over {\sin \theta \cos \theta }} = 1 + \cos ec\theta \sec \theta$
= R.H.S.

(iv) R.H.S. = ${{{{\sin }^2}A} \over {1 - \cos A}} = {{1 - {{\cos }^2}A} \over {1 - \cos A}}$
$[Since,{\sin ^2}A = 1 - {\cos ^2}A]$
= ${{\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)} \over {1 - \cos A}} = 1 + \cos A$
$[Since,{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]$
= $1 + {1 \over {\sec A}} = {{1 + \sec A} \over {\sec A}} = L.H.S.$

(v) L.H.S. = ${{\cos A - \sin A + 1} \over {\cos A + \sin A - 1}} = {{{{\cos A - \sin A + 1} \over {\sin A}}} \over {{{\cos A + \sin A - 1} \over {\sin A}}}}$
= ${{\cos A - 1 + \cos ecA} \over {\cos A + 1 - \cos ecA}}$
$[Since,1 + {\cot ^2}A = \cos e{c^2}A]$
$= {{\cot A + \cos ecA - \left( {\cos e{c^2}A - {{\cot }^2}A} \right)} \over {\cot A - \cos ecA + 1}}$
= ${{\cot A + \cos ecA - \left( {\cos ecA + \cot A} \right)\left( {\cos ecA - \cot A} \right)} \over {\cot A - \cos ecA + 1}}$
$[Since{A^2} - {B^2} = \left( {A + B} \right)\left( {A - B} \right)]$
Taking common(cosecA + cotA)
= ${{\left( {\cos ecA + \cot A} \right)\left( {1 - \cos ecA + \cot } \right)} \over {\left( {\cot A - \cos ecA + 1} \right)}}$
= cosec A + cot A
= R.H.S.

(vi) We have,
L.H.S. = $\sqrt {{{1 + \sin A} \over {1 - \sin A}}} = \sqrt {{{1 + \sin A} \over {1 - \sin A}} \times {{1 + \sin A} \over {1 + \sin A}}}$
[Multiplying and dividing by ] $\sqrt {1 + \sin A}$
= $\sqrt {{{{{\left( {1 + \sin A} \right)}^2}} \over {1 - {{\sin }^2}A}}} = \sqrt {{{{{\left( {1 + \sin A} \right)}^2}} \over {{{\cos }^2}A}}}$ $[Since,\,{\sin ^2}A + {\cos ^2}A = 1]$
= ${\sqrt {\left( {{{1 + \sin A} \over {\cos A}}} \right)} ^2} = {{1 + \sin A} \over {\cos A}}$
= ${1 \over {\cos A}} + {{\sin A} \over {\cos A}} = \sec A + \tan A$
= R.H.S. $\left[ {Since,\,\tan A = {{\sin A} \over {\cos A}}} \right]$

(vii) We, have,
L.H.S. = ${{\sin \theta - 2{{\sin }^3}\theta } \over {2{{\cos }^3}\theta - \cos \theta }} = {{\sin \theta (1 - 2{{\sin }^2}\theta )} \over {\cos \theta \left( {2{{\cos }^2}\theta - 1} \right)}}$
= $\tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {2\left( {1 - {{\sin }^2}\theta } \right) - 1}}} \right]$
= $\tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {2 - 2{{\sin }^2}\theta - 1}}} \right]$
= $\tan \theta \left[ {{{1 - 2{{\sin }^2}\theta } \over {1 - 2{{\sin }^2}\theta }}} \right] = \tan \theta \times 1$
= $\tan \theta$ = R.H.S.

(viii) We have,
L.H.S. = ${\left( {\sin A + \cos ecA} \right)^2} + {\left( {\cos A + \sec A} \right)^2}$
= $({\sin ^2}A + \cos e{c^2}A + 2\sin A\cos ecA) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2\cos A\sec A} \right)$
= $\left( {{{\sin }^2}A + \cos e{c^2}A + 2\sin A.{1 \over {\sin A}}} \right) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2\cos A.{1 \over {\cos A}}} \right)$
= $\left( {{{\sin }^2}A + \cos e{c^2}A + 2} \right) + \left( {{{\cos }^2}A + {{\sec }^2}A + 2} \right)$
= ${\sin ^2}A + {\cos ^2}A + \cos e{c^2}A + {\sec ^2}A + 4$ $[Since,\,\,{\sin ^2}A + {\cos ^2}A = 1]$
= $1 + \left( {1 + {{\cot }^2}A} \right) + \left( {1 + {{\tan }^2}A} \right) + 4$
= $7 + {\tan ^2}A + {\cot ^2}A$
= $[Since,\,\cos e{c^2}A = 1 + {\cot ^2}A\,\,and\,{\sec ^2}A = 1\, + \,{\tan ^2}A]$
= R.H.S.

(ix) We have,
L.H.S. = (cosec A – sinA) (secA – cosA)
= $\left( {{1 \over {\sin A}} - \sin A} \right)\left( {{1 \over {\cos A}} - \cos A} \right)$
= $\left( {{{1 - {{\sin }^2}A} \over {\sin A}}} \right)\left( {{{1 - {{\cos }^2}A} \over {\cos A}}} \right)$
= ${{{{\cos }^2}A} \over {\sin A}} \times {{{{\sin }^2}A} \over {\cos A}}$
= sinA cosA
= ${{\sin A\cos A} \over {{{\sin }^2}A + {{\cos }^2}A}}$ $[Since,{\sin ^2}A + {\cos ^2}A = 1]$
Dividing Numerator and Denominator by sinA cosA
${{{{\sin A\cos A} \over {\sin A\cos A}}} \over {{{{{\sin }^2}A} \over {\sin A\cos A}} + {{{{\cos }^2}A} \over {\sin A\cos A}}}}$
= ${1 \over {{{\sin A} \over {\cos A}} + {{\cos A} \over {\sin A}}}}$
= ${1 \over {\tan A + \cot A}}$ = R.H.S.

(x) We have,
L.H.S. = $\left( {{{1 + {{\tan }^2}A} \over {1 + {{\cot }^2}A}}} \right) = {{{{\sec }^2}A} \over {\cos e{c^2}A}}$
= ${1 \over {{{\cos }^2}A}} \times {{{{\sin }^2}A} \over 1} = {\tan ^2}A$
R.H.S. = ${\left( {{{1 - \tan A} \over {1 - \cot A}}} \right)^2} = {\left( {{{1 - \tan A} \over {1 - {1 \over {\tan A}}}}} \right)^2}$
= ${\left( {{{1 - \tan A} \over {{{\tan A - 1} \over {\tan A}}}}} \right)^2} = {\left( { - \tan A} \right)^2} = {\tan ^2}A$
Therefore, L.H.S. = R.H.S.

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