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ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios
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From the figure (1) given below, find the value of sec …
(b) From the figure (2) given below, find the values of
(i) sin x
(ii) cot x
\text { (iii) } \cot ^{2} x-\operatorname{cosec}^{2} x
(iv) sec y
(v) tan ……
(a) From the figure, Sec θ = AB / BD
\text { But in } \Delta \mathrm{ADC}, \angle \mathrm{D}=90^{\circ}
\begin{aligned} &\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2} \text { (Pythagoras Theorem }\\ &(13)^{2}=A D^{2}+25\\ &A D^{2}=169-25\\ &=144\\ &=(12)^{2}\\ &A D=12\\ &\text { (in right } \Delta \mathrm{ABD})\\ &A B^{2}=A D^{2}+B D^{2}\\ &=(12)^{2}+(16)^{2}\\ &=144+256\\ &=400\\ &=(20)^{2} \end{aligned}
AB = 20
Now, Sec θ = AB / BD
= 20/16
= 5/4
(b) let given ∆ABC
BD = 3, AC = 12, AD = 4
In the right-angled ∆ABD
By Pythagoras theorem
\begin{array}{l} A B^{2}=A D^{2}+B D^{2} \\ A B^{2}=(4)^{2}+(3)^{2} \\ A B^{2}=16+9 \end{array}
\begin{array}{l} A B^{2}=25 \\ A B^{2}=(5)^{2} \\ A B=5 \end{array}
In right angled triangle ACD
By Pythagoras theorem,
\begin{array}{l} A C^{2}=A D^{2}+C D^{2} \\ C D^{2}=A C^{2}-A D^{2} \\ C D^{2}=(12)^{2}-(4)^{2} \\ C D^{2}=144-16 \\ C D^{2}=128 \end{array}
CD = √128
CD = √64 × 2 CD
= 8√2
(i) sin x = perpendicular/Hypotenuse
= AD/AB
= 4/5
(ii) cot x = Base/Perpendicular
= BD/AD
= ¾
(iii) cot x = Base/ Perpendicular
=BD/AD
= 3/4
cosec x = Hypotenuse / Perpendicular
AB/BD
= 5/4
\begin{array}{l} \cot ^{2} x-\operatorname{cosec}^{2} x \\ =(3 / 4)^{2}-(5 / 4)^{2} \end{array}
= 9/16 – 25/16
(9 -25)/16
= -16/16
= -1
Perpendicular = Hypotenuse/Base (in right angled ∆ACD)
= AD/CD
= 12/(8 √2)
= 3/(2 √2)
Cot y = Base/ Hypotenuse
= AD/CD
= 4/8 √ 2
= 1/2 √2
Cot y = Base / Hypotenuse ((in right angled ∆ACD)
= CD/AC
= 8√2/12
= 2√/3
\begin{array}{l} \text { Now } \tan ^{2} y-1 / \cot ^{2} y \\ =(1 / 2 \sqrt{2})^{2}-1 /(2 \sqrt{2} / 3)^{2} \end{array}
= ¼ × - ¼ × 2
= (1/8) – (9/8)
= (1-9)/8
= -8/8
= -1
\tan ^{2} y-1 / \cot ^{2} y=-1
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