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From the Following Figure, Find the Values of : Cos A Cosec a Tan2a - Sec2a Sin C Sec C Cot2 C - 1 / Sin^2 C - Mathematics

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प्रश्न

From the following figure, find the values of
(i) cos A 
(ii) cosec A
(iii) tan2A - sec2
(iv) sin C
(v) sec C 
(vi) cot2 C - ` 1 / sin^2 "c"`

योग

उत्तर

Consider the diagram as

Given angle  ADB = 90° and BDC = 90°
⇒ AB2 = AD2 + BD2         ...( AB is hypotenuse in ΔABD )

⇒ AB2 = 32 + 42

∴ AB2 = 9 + 16 = 25 and AB = 5

⇒ BC2 = BD2 + DC2       ...( BC is hypotenuse in ΔBDC )

⇒ DC2 = 122 - 42

∴ DC2 = 144 - 16 = 128 aand DC = 8`sqrt2` 

(i) cos A = `"base"/"hypotenuse" = "AD"/"AB" =3/5`

(ii) cosec A = `"hypotenuse"/"perpendicular" = "AB"/"BD" =5/4`

(iii) tan A = `"perpendicular"/"base" = "BD"/"AD" =4/3`

sec A = `"hypotenuse"/"base" = "AB"/"AD" =5/3`

tan2 A - sec2 A = `(4/3)^2 - (5/3)^2`

= `16/9 - 25/9`

= `( – 9)/(9)`

= – 1

(iv) sin C = `"perpendicular"/"hypotenuse" = "BD"/"BC" = 4/12 = 1/3`

(v) sec C = `"hypotenuse"/"base" = "BC"/"DC" = 12/(8sqrt2) = 3/(2sqrt2) = (3sqrt2)/4`

(vi) cot C = `"base"/"perpendicular" = "DC"/"BD" = (8sqrt2)/4 = 2sqrt2`

sin C = `"perpendicular"/"hypotenuse" = "BD"/"BC" = 4/12 = 1/3`

cot2 C –  ` 1/sin^2C = (2sqrt2)^2 - 1/(1/3)^2`

= 8 - 9
=  – 1                        

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अध्याय 22: Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] - Exercise 22 (A) [पृष्ठ २७९]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 22 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]
Exercise 22 (A) | Q 3 | पृष्ठ २७९
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