ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios

Question 3 Trigonometric Ratios Exercise 17

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

= 4/5

(ii) cot x = Base/Perpendicular

= ¾

(iii) cot x = Base/ Perpendicular

= 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)

= 12/(8 √2)

= 3/(2 √2)

Cot y = Base/ Hypotenuse

= 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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