{"id":7438,"date":"2024-01-21T10:49:50","date_gmt":"2024-01-21T05:19:50","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=7438"},"modified":"2024-01-23T09:36:21","modified_gmt":"2024-01-23T04:06:21","slug":"ts-inter-1st-year-maths-1a-solutions-chapter-8-ex-8a","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-solutions-chapter-8-ex-8a\/","title":{"rendered":"TS Inter 1st Year Maths 1A Solutions Chapter 8 Inverse Trigonometric Functions Ex 8(a)"},"content":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions<\/a> Chapter 8 Inverse Trigonometric Functions Ex 8(a) to find a better approach to solving the problems.<\/p>\n

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a)<\/h2>\n

I.
\n1.
\nEvaluate the following:
\n(i) Sin-1<\/sup> \\(\\left(-\\frac{\\sqrt{3}}{2}\\right)\\)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) Cos-1<\/sup>\\(\\left(\\frac{1}{\\sqrt{2}}\\right)\\)
\nAnswer:
\ncos-1<\/sup>\\(\\frac{1}{\\sqrt{2}}\\) = cos-1<\/sup> (cos\\(\\frac{\\pi}{4}\\)) = \\(\\frac{\\pi}{4}\\)
\n(\u2235 cos-1<\/sup> (cos \u03b8) = \u03b8 if \u03b8 \u2208 [0, \u03c0])<\/p>\n

(iii) Sec-1<\/sup> (- \u221a2)
\nAnswer:
\n= \u03c0 – sec-1<\/sup> \u221a2 [\u2235 x \u2208 ( – \u221e, – 1] \u222a [1, \u221e)]
\n= \u03c0 – sec-1<\/sup> (sec \\(\\frac{\\pi}{4}\\)) = \u03c0 – \\(\\frac{\\pi}{4}\\) = \\(\\frac{3 \\pi}{4}\\)
\n[\u2235 sec-1<\/sup> (sec \u03b8) = \u03b8 dor \u03b8 \u2208 \\(\\left(0, \\frac{\\pi}{2}\\right)\\) \u222a \\(\\left(\\frac{\\pi}{2}, \\pi\\right)\\)]<\/p>\n

(iv) Cot-1<\/sup> (- \u221a3)
\nAnswer:
\nCot-1<\/sup> (- \u221a3) = \u03c0 – cot-1<\/sup> (\u221a3)
\n= \u03c0 – cot-1<\/sup> (cot\\(\\frac{\\pi}{6}\\))
\n= \u03c0 – \\(\\frac{\\pi}{6}\\) = \\(\\frac{5 \\pi}{6}\\)
\n(\u2235 cot-1<\/sup> (- x) = \u03c0 – cot-1<\/sup>x for any x \u2208 R)<\/p>\n

(v) sin (\\(\\frac{\\pi}{3}\\) – sin-1<\/sup>\\(\\left(-\\frac{1}{2}\\right)\\))
\nAnswer:
\n\"TS<\/p>\n

(vi) Sin-1<\/sup> \\(\\left(\\sin \\left(\\frac{5 \\pi}{6}\\right)\\right)\\)
\nAnswer:
\nsin-1<\/sup> \\(\\left(\\sin \\left(\\pi-\\frac{\\pi}{6}\\right)\\right)\\)
\n= sin-1<\/sup> (sin\\(\\frac{\\pi}{6}\\)) = \\(\\frac{\\pi}{6}\\)<\/p>\n

(vii) Cos-1<\/sup>\\(\\left(\\cos \\frac{5 \\pi}{4}\\right)\\)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the value of
\n(i) Sin \\(\\left(\\cos ^{-1}\\left(\\frac{3}{5}\\right)\\right)\\)
\nAnswer:
\n\"TS<\/p>\n

(ii) Tan (cosec-1<\/sup> \\(\\left(\\frac{65}{63}\\right)\\))
\nAnswer:
\n\"TS<\/p>\n

(iii) Sin (2 sin-1<\/sup> \\(\\frac{4}{5}\\))
\nAnswer:
\n\"TS<\/p>\n

(iv) Sin-1<\/sup>\\(\\left(\\sin \\frac{33 \\pi}{7}\\right)\\)
\nAnswer:
\n\"TS<\/p>\n

(v) Cos-1<\/sup> (cos \\(\\frac{17 \\pi}{6}\\))
\nAnswer:
\n\"TS<\/p>\n

Question 3.
\nSimplify each of the following.
\n(i) Tan-1<\/sup>\\(\\left(\\frac{\\sin x}{1+\\cos x}\\right)\\)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) tan-1<\/sup> (sec x + tan x)
\nAnswer:
\ntan-1<\/sup> (sec x + tan x)
\n\"TS<\/p>\n

(iii) Tan-1<\/sup> \\(\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\)
\nAnswer:
\n\"TS<\/p>\n

(iv) Sin-1<\/sup> (2cos2<\/sup>\u03b8 – 1) + Cos-1<\/sup> (1 – 2 sin2<\/sup> \u03b8)
\nAnswer:
\nsin-1<\/sup> (2cos2<\/sup>\u03b8 – 1) + cos-1<\/sup> (1 – 2 sin2<\/sup> \u03b8)
\n= sin-1<\/sup> (cos2\u03b8) + cos-1<\/sup> (cos 2\u03b8)
\n= sin-1<\/sup> (sin(\\(\\frac{\\pi}{2}\\) – 2\u03b8) + cos-1<\/sup> (cos 2\u03b8))
\n= \\(\\frac{\\pi}{2}\\) – 2\u03b8 + 2\u03b8= \\(\\frac{\\pi}{2}\\)<\/p>\n

(v) tan-1<\/sup> (x + \\(\\sqrt{1+x^2}\\)) = ; x \u2208 R
\nAnswer:
\nLet x = tan \u03b8 then tan-1<\/sup> (x + \\(\\sqrt{1+x^2}\\)) = tan-1<\/sup> (tan \u03b8 + sec \u03b8)
\n\"TS<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nProve that
\n(i) Sin-1<\/sup>\\(\\frac{3}{5}\\) + sin-1<\/sup>\\(\\frac{8}{17}\\) = Cos-1<\/sup>\\(\\left(\\frac{36}{85}\\right)\\) (May 2022)
\nAnswer:
\nUse the property that If
\nx, y \u2208 [0, 1] and x2<\/sup> + y2<\/sup> < 1
\n(Here x = \\(\\frac{3}{5}\\) and y = \\(\\frac{8}{17}\\)) then
\nsin-1<\/sup> x + sin-1<\/sup> y
\n= cos-1<\/sup> (\\(\\sqrt{1-x^2} \\sqrt{1-y^2}\\) – xy)
\nAs an alternative preferably use
\nsin-1<\/sup>\\(\\left(\\frac{3}{5}\\right)\\) = A and sin-1<\/sup>\\(\\left(\\frac{8}{17}\\right)\\) = B
\n\u2234 sinA = \\(\\frac{3}{5}\\) and sinB = \\(\\frac{8}{17}\\)
\n\u2234 cosA = \\(\\frac{4}{5}\\) andcosB = \\(\\frac{15}{17}\\)
\nNow A + B \u2208 (0, \u03c0) and cos (A + B)
\n= cos A cos B – sin A sin B
\n\"TS<\/p>\n

(ii) Sin-1<\/sup>\\(\\left(\\frac{3}{5}\\right)\\) + Cos-1<\/sup>\\(\\left(\\frac{12}{13}\\right)\\) = Cos-1<\/sup>\\(\\left(\\frac{33}{65}\\right)\\)
\nAnswer:
\nLet sin-1<\/sup>\\(\\left(\\frac{3}{5}\\right)\\) = A and cos-1<\/sup>\\(\\left(\\frac{12}{13}\\right)\\) then A + B \u2208 (0, \u03c0)
\n\u2234 sinA = \\(\\frac{3}{5}\\) and cos B = \\(\\frac{12}{13}\\)
\n\u2234 cosA = \\(\\frac{4}{5}\\) and SinB = \\(\\frac{5}{13}\\)
\nConsider cos (A + B) = cos A cos B – sin A sin B
\n\"TS<\/p>\n

(iii) tan (cot-1<\/sup>9 + Cosec-1<\/sup>\\(\\frac{\\sqrt{41}}{4}\\)) = 1
\nAnswer:
\nLet cot-1<\/sup>9 = A and cosec-1<\/sup>\\(\\frac{\\sqrt{41}}{4}\\) = B
\nthen cot A = 9 and cosec B = \\(\\frac{\\sqrt{41}}{4}\\)
\n\u2234 tan A = \\(\\frac{1}{9}\\) and cot2<\/sup>B = cosec2<\/sup>B – 1 = \\(\\frac{41}{16}\\) – 1 = \\(\\frac{25}{16}\\)
\n\u21d2 cot B = \\(\\frac{41}{16}\\) – 1 \u21d2 tan B = \\(\\frac{25}{16}\\)
\n\u21d2 cot B = \\(\\frac{5}{4}\\) \u21d2 tan B = \\(\\frac{4}{5}\\)
\n\"TS<\/p>\n

\"TS<\/p>\n

(iv) Cos-1<\/sup>\\(\\left(\\frac{4}{5}\\right)\\) + sin-1<\/sup>\\(\\left(\\frac{3}{\\sqrt{34}}\\right)\\) = tan-1<\/sup>\\(\\left(\\frac{27}{11}\\right)\\)
\nAnswer:
\n\"TS<\/p>\n

Question 2.
\nFind the value of
\n(i) sin (Cos-1<\/sup>\\(\\frac{3}{5}\\) + Cos-1<\/sup>\\(\\frac{12}{13}\\))
\nAnswer:
\nLet Cos-1<\/sup>\\(\\frac{3}{5}\\) = A and Cos-1<\/sup>\\(\\frac{12}{13}\\) = B
\nthen cos A = \\(\\frac{3}{5}\\) and cos B = \\(\\frac{12}{13}\\)
\n\u2234 sin A = \\(\\frac{4}{5}\\) and sinB = \\(\\frac{5}{13}\\)
\n\u2234 sin(A + B) = sin A cos B + cos A sin B
\n\"TS<\/p>\n

(ii) tan (sin-1<\/sup>\\(\\left(\\frac{3}{5}\\right)\\) + cos-1<\/sup>\\(\\left(\\frac{5}{\\sqrt{34}}\\right)\\))
\nAnswer:
\n\"TS<\/p>\n

(iii) cos (Sin-1<\/sup> \\(\\frac{3}{5}\\) + Sin-1<\/sup> \\(\\frac{5}{13}\\))
\nAnswer:
\nLet Sin-1<\/sup>\\(\\left(\\frac{3}{5}\\right)\\) = A and Sin-1<\/sup>\\(\\left(\\frac{5}{13}\\right)\\) = B
\nthen sin A = \\(\\frac{3}{5}\\) and sin B = \\(\\frac{5}{13}\\)
\n\u2234 cosA = \\(\\frac{4}{5}\\) and cos B = \\(\\frac{12}{13}\\)
\n\u2234 cos (A + B) = cos A cos B – sin A sin B
\n= \\(\\left(\\frac{4}{5}\\right)\\left(\\frac{12}{13}\\right)\\) – \\(\\left(\\frac{3}{5}\\right)\\left(\\frac{5}{13}\\right)\\) = \\(\\frac{33}{65}\\)<\/p>\n

\"TS<\/p>\n

Question 3.
\nProve that
\n(i) cos (2 Tan-1<\/sup>\\(\\frac{1}{7}\\)) = sin (2 Tan-1<\/sup>\\(\\frac{3}{4}\\))
\nAnswer:
\n\"TS<\/p>\n

(ii) cos {2[Tan-1<\/sup>\\(\\left(\\frac{1}{4}\\right)\\) + Tan-1<\/sup>\\(\\left(\\frac{2}{9}\\right)\\)]} = \\(\\frac{3}{5}\\)
\nAnswer:
\n\"TS<\/p>\n

Question 4.
\nProve that
\n(i) Tan-1<\/sup>\\(\\frac{1}{7}\\) + Tan-1<\/sup>\\(\\frac{1}{13}\\) – Tan-1<\/sup>\\(\\frac{2}{9}\\) = 0
\nAnswer:
\nL.H.S = (tan-1<\/sup>\\(\\frac{1}{7}\\) + tan-1<\/sup>\\(\\frac{1}{13}\\)) – tan-1<\/sup>\\(\\frac{2}{9}\\)
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) Tan-1<\/sup>\\(\\frac{1}{2}\\) + Tan-1<\/sup>\\(\\frac{1}{5}\\) – Tan-1<\/sup>\\(\\frac{1}{8}\\) = \\(\\frac{\\pi}{4}\\) (March 2015-A.P.) (March 2011, May 2006)
\nAnswer:
\n\"TS<\/p>\n

(iii) Tan-1<\/sup>\\(\\frac{3}{4}\\) + Tan-1<\/sup>\\(\\frac{3}{5}\\) – Tan-1<\/sup>\\(\\frac{8}{19}\\) = \\(\\frac{\\pi}{4}\\)
\nAnswer:
\n\"TS<\/p>\n

(iv) Tan-1<\/sup>\\(\\frac{1}{7}\\) + Tan-1<\/sup>\\(\\frac{1}{8}\\) = Cot-1<\/sup>\\(\\frac{201}{43}\\) + Cot-1<\/sup> 18
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nShow that
\n(i) sec2<\/sup> (Tan-1<\/sup>2) + cosec2<\/sup> (Cot-1<\/sup>2) = 10
\nAnswer:
\nLet \u03b1 = tan-1<\/sup>2 tan \u03b1 = 2
\nsec2<\/sup> \u03b1 = 1 + tan2<\/sup>\u03b1 = 1 + 4 = 5
\nLet \u03b2 = cot-1<\/sup> 2 \u21d2 cot \u03b2 = 2
\n\u2234 cosec2<\/sup> \u03b2 = 1 + cot2<\/sup> \u03b2 = 1 + 4 = 5
\n\u2234 sec2<\/sup> (tan-1<\/sup> 2) + cosec2<\/sup> (cot-1<\/sup>2)
\n= sec2<\/sup>\u03b1 + cosec2<\/sup>\u03b2 = 5 + 5 = 10<\/p>\n

(ii) Find the value of tan(Cos-1<\/sup>\\(\\frac{4}{5}\\) + Tan-1<\/sup>\\(\\frac{2}{3}\\)). (March 2012)
\nAnswer:
\n\"TS<\/p>\n

(iii) If sin-1<\/sup>x – Cos-1<\/sup> = \\(\\frac{\\pi}{6}\\), then find x.
\nAnswer:
\nLet \u03b1 = sin-1<\/sup>x \u21d2 sin \u03b1 = x
\nand \u03b2 = cos-1<\/sup>x \u21d2 cos \u03b2 = x
\n\"TS<\/p>\n

III.
\nQuestion 1.
\nProve that (May 2014, Mar. 14)
\n(i) 2 sin-1<\/sup>\\(\\frac{3}{5}\\) – cos-1<\/sup>\\(\\frac{5}{13}\\) = cos-1<\/sup>\\(\\left(\\frac{323}{325}\\right)\\)
\nAnswer:
\nLet \u03b1 = sin-1<\/sup>\\(\\left(\\frac{3}{5}\\right)\\) and \u03b2 = cos-1<\/sup>\\(\\left(\\frac{5}{13}\\right)\\)
\n\u21d2 sin \u03b1 = \\(\\frac{3}{5}\\) and cos \u03b2 = \\(\\frac{5}{13}\\)
\n\u2234 LHS = 2\u03b1 – \u03b2
\nConsider cos(2\u03b1 – \u03b2)
\n= cos 2\u03b1 cos \u03b2 + sin 2\u03b1 sin \u03b2
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) Sin-1<\/sup>\\(\\left(\\frac{4}{5}\\right)\\) + 2 Tan-1<\/sup> \\(\\left(\\frac{1}{3}\\right)\\) = \\(\\frac{\\pi}{2}\\) (March 2015-T.S)
\nAnswer:
\n\"TS<\/p>\n

(iii) 4 tan-1<\/sup>\\(\\left(\\frac{1}{5}\\right)\\) + Tan-1<\/sup>\\(\\left(\\frac{1}{99}\\right)\\) – Tan-1<\/sup>\\(\\left(\\frac{1}{70}\\right)\\) = \\(\\frac{\\pi}{4}\\)
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\n(i) If \u03b1 = Tan-1<\/sup>\\(\\left(\\frac{\\sqrt{1+x^2}-\\sqrt{1-x^2}}{\\sqrt{1+x^2}+\\sqrt{1-x^2}}\\right)\\), then prove that x2<\/sup> = sin 2 \u03b1.
\nAnswer:
\nLet x2<\/sup> = \u03b8
\n\"TS<\/p>\n

(ii) Prove that tan\\(\\left\\{2 {Tan}^{-1}\\left(\\frac{\\sqrt{1+x^2}-1}{x}\\right)\\right\\}\\) = x.
\nAnswer:
\nLet x = tan \u03b8
\n\"TS<\/p>\n

(iii) Prove that
\nsin\\(\\left[{Cot}^{-1} \\frac{2 x}{1-x^2}+{Cos}^{-1}\\left(\\frac{1-x^2}{1+x^2}\\right)\\right]\\) = 1.
\nAnswer:
\nLet x = tan \u03b8 then
\n\"TS<\/p>\n

\"TS<\/p>\n

(iv) prove that tan \\(\\left\\{\\frac{\\pi}{4}+\\frac{1}{2} \\cos ^{-1}\\left(\\frac{a}{b}\\right)\\right\\}\\) + tan \\(\\left\\{\\frac{\\pi}{4}-\\frac{1}{2} \\cos ^{-1}\\left(\\frac{a}{b}\\right)\\right\\}\\) = \\(\\frac{2 b}{a}\\)
\nAnswer:
\n\"TS<\/p>\n

Question 3.
\n(i) If Cos-1<\/sup> p + Cos-1<\/sup> q + Cos-1<\/sup> r = \u03c0, then prove that p2<\/sup> + q2<\/sup> + r2<\/sup> + 2pqr = 1. (June 2004, 2006, 2005)
\nAnswer:
\nLet cos-1<\/sup>p = A, cos-1<\/sup> q = B, cos-1<\/sup>r = c
\n\u21d2 cos A = p, cos B = q, and cos C = \u03c0
\nthen given A + B + C = \u03c0
\nTo show that p2<\/sup> + q2<\/sup> + r2<\/sup> = 1 – 2 pqr
\ni.e., cos2<\/sup>A + cos2<\/sup> B + cos2<\/sup>C = 1 – 2 cos A cos B cos C
\n\u2234 cos2<\/sup> A + cos2<\/sup> B + cos2<\/sup> C = cos2<\/sup> A + cos2<\/sup> B + 1 – sin2<\/sup> C
\n= 1 + cos2<\/sup> A + cos (B +C) cos (B – C)
\n= 1 + cos2<\/sup> A – cos A cos (B – C)
\n= 1 + cos A [ cos A – cos (B – C)]
\n= 1 + cos A [ – cos (B + c) – cos (B – C)]
\n= 1 – cos A [ cos (B + C) + cos (B – C)]
\n= 1 – 2 cos A cosB cos C = 1 – 2 pqr<\/p>\n

(ii) If sin-1<\/sup>\\(\\left(\\frac{2 p}{1+p^2}\\right)\\) – cos-1<\/sup>\\(\\left(\\frac{1-q^2}{1+q^2}\\right)\\) = Tan-1<\/sup>\\(\\left(\\frac{2 x}{1-x^2}\\right)\\), then prove that x = \\(\\).
\nAnswer:
\nLet p = tan A, q = tan B and x = tan C
\n\"TS
\n\u21d2 sin-1<\/sup> (sin 2A) – cos-1<\/sup> (cos 2B) = tan-1<\/sup> (tan 2C)
\n\u21d2 2A – 2B = 2C
\n\u2234 tan C = \\(\\frac{\\tan A \\tan B}{1+\\tan A \\tan B}\\) = \\(\\frac{p-q}{1+p q}\\)
\n\u21d2 x = \\(\\frac{p-q}{1+p q}\\)<\/p>\n

\"TS<\/p>\n

(iii) If a, b, c are distinct non – zero real numbers having the same sign prove that
\ncot-1<\/sup>\\(\\left(\\frac{a b+1}{a-b}\\right)\\) + cot-1<\/sup>\\(\\left(\\frac{b c+1}{b-c}\\right)\\) + cot-1<\/sup>\\(\\left(\\frac{c a+1}{c-a}\\right)\\) = \u03c0 or 2\u03c0.
\nAnswer:
\nLet have (a – b) + (b – c) + (c – a) = 0 and (a – b), (b – c), (c – a) need not have the same sign. Then either two of these numbers may be positive and one negative or two may be negative and one will be positive. Suppose (a – b) (b – c) are both positive and (c – a) is negative.
\n\"TS<\/p>\n

(iv) If sin-1<\/sup>x + sin-1<\/sup>y + sin-1<\/sup>z = \u03c0, then prove that
\nx \\(\\sqrt{1-x^2}\\) + y\\(\\sqrt{1-y^2}\\) + y\\(\\sqrt{1-z^2}\\) = 2xyz (March 2006, May 2005)
\nAnswer:
\nLet sin-1<\/sup> x = A, sin-1<\/sup> y = B and sin-1<\/sup> z = c
\nThen sin A = x, sin B = y and sin c = z
\nAlso A + B + C = \u03c0
\nLHS
\n= x\\(\\sqrt{1-x^2}\\) + y\\(\\sqrt{1-y^2}\\) + z\\(\\sqrt{1-z^2}\\)
\n= sin A cos A + sin B cos B + smC cos C
\n= \\(\\frac{1}{2}\\) [sin 2A + sin 2B + sin 2C]
\n= \\(\\frac{1}{2}\\) [2.sin(A + B) cos (A – B) + 2 sin C cos C]
\n= \\(\\frac{1}{2}\\) [2 sin C cos (A – B) + 2 sin C cos C]
\n= sin C [cos (A – B) + cos C]
\n= sin C [cos (A – B) – cos (A + B)]
\n= 2 sin C sin A sin B = 2 sin A sin B sin C
\n= 2xyz = RHS<\/p>\n

(v) (a) If Tan-1<\/sup> x + Tan-1<\/sup> y + Tan-1<\/sup> z = \u03c0, then prove that x + y + z = xyz. (March 2003)
\nAnswer:
\nLet tan-1<\/sup> x = A, tan-1<\/sup> y = B and tan-1<\/sup> z = c then A + B + C = \u03c0 and tan A = x, tan B = y, tan C = z.
\n\u2234 A + B = (\u03c0 – C)
\n\u21d2 tan (A + B) = tan(\u03c0 – C) = – tan C
\n\u21d2 \\(\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}\\) = – tan C
\n\u21d2 tan A + tan B = – tan C + tan A tan B tan C
\n\u21d2 tan A + tan B + tan C = tan A tan B tan C
\n\u21d2 x + y + z = XYZ.<\/p>\n

\"TS<\/p>\n

(b) If Tan-1<\/sup>x + Tan-1<\/sup>y + Tan-1<\/sup>z = \\(\\) then prove that xy + yz + zx = 1. (March 2001, June 2004)
\nAnswer:
\nLet tan-1<\/sup> x = A, tan-1<\/sup> y = B, tan-1<\/sup> z = C
\nthen A + B + C = \\(\\frac{\\pi}{2}\\)
\n\u2234 A + B + C = \\(\\frac{\\pi}{2}\\) – C
\n\u21d2 tan (A + B) = tan (\\(\\frac{\\pi}{2}\\) – C) = cot C = \\(\\frac{1}{\\tan C}\\)
\n\u2234 \\(\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}\\) = \\(\\frac{1}{\\tan C}\\)
\n\u21d2 tan A + tan B + tan B tan C + tan C tan A = 1
\n\u21d2 xy + yz + zx = 1<\/p>\n

Question 4.
\nSolve the following equation for x:
\n(i) Tan-1<\/sup>\\(\\left(\\frac{x-1}{x-2}\\right)\\) + Tan-1<\/sup>\\(\\left(\\frac{x+1}{x+2}\\right)\\) = \\(\\frac{\\pi}{4}\\)
\nAnswer:
\n\"TS<\/p>\n

(ii) Tan-1<\/sup>\\(\\left(\\frac{1}{2 x+1}\\right)\\) + Tan-1<\/sup>\\(\\left(\\frac{1}{4 x+1}\\right)\\) = Tan-1<\/sup>\\(\\left(\\frac{2}{x^2}\\right)\\)
\nAnswer:
\n\"TS
\n\u21d2 x2<\/sup>(3x + 1) = 2x(4x + 3)
\n\u21d2 x = 0 or 6x2<\/sup> – 14x – 12 = 0
\n\u21d2 x = 0 or 3x2<\/sup> – 7x – 6 = 0
\n\u21d2 x = 0 or (3x2<\/sup> – 9x + 2x – 6) = 0
\n\u21d2 x = 0 or 3x(x – 3) + 2(x – 3) = 0
\n\u21d2 x = 0 or (x – 3) (3x – 2) = 0
\n\u21d2 x = 0 (or) x = 3 or x = – \\(\\frac{2}{3}\\)<\/p>\n

(iii) 3 sin-1<\/sup>\\(\\left(\\frac{2 x}{1+x^2}\\right)\\) – 4 cos \\(\\left(\\frac{1-x^2}{1+x^2}\\right)\\) + 2 Tan-1<\/sup>\\(\\left(\\frac{2 x}{1-x^2}\\right)\\) = \\(\\frac{\\pi}{3}\\)
\nAnswer:
\nLet x = tan\u03b8 then
\n\"TS<\/p>\n

\"TS<\/p>\n

(iv) sin-1<\/sup> (1 – x) – 2 sin-1<\/sup> x = \\(\\frac{\\pi}{2}\\)
\nAnswer:
\nLet sin-1<\/sup> (1 – x) = \u03b1 and sin-1<\/sup> x = \u03b2 then sin \u03b1 = 1 – x and sin \u03b2 = x
\n\"TS
\nHence x = 0 is the only solution for the given equation.<\/p>\n

Question 5.
\nSolve the following equations.
\n(i) Cot-1<\/sup> \\(\\left(\\frac{1+x}{1-x}\\right)\\) = \\(\\frac{1}{2}\\) Cot-1<\/sup>\\(\\left(\\frac{1}{x}\\right)\\), x > 0 and x \u2260 1.
\nAnswer:
\n\"TS<\/p>\n

(ii) Tan[Cos-1<\/sup>\\(\\left(\\frac{1}{\\mathbf{x}}\\right)\\)] = Sin[Cot-1<\/sup>\\(\\left(\\frac{1}{2}\\right)\\)]; x \u2260 0.
\nAnswer:
\n\"TS<\/p>\n

\"TS<\/p>\n

(iii) Cos-1<\/sup> x + Sin-1<\/sup>\\(\\left(\\frac{x}{2}\\right)\\) = \\(\\frac{\\pi}{6}\\)
\nAnswer:
\n\"TS
\n\u21d2 x4<\/sup> – 5x2<\/sup> + 4 = x4<\/sup> – 2x2<\/sup> + 1
\n\u21d2 3x2<\/sup> – 3 = 0 \u21d2 3x2<\/sup> = 3 \u21d2 x = \u00b1 1
\nWhen x = – 1 then cos-1<\/sup> (- 1) + sin-1<\/sup>\\(\\left(-\\frac{1}{2}\\right)\\)
\n= \u03c0 – \\(\\frac{\\pi}{6}\\) = \\(\\frac{5 \\pi}{6}\\) \u2260 \\(\\frac{\\pi}{6}\\)
\n\u2234 x = – 1 is not admissible. Hence x = 1<\/p>\n

(iv) Cos-1<\/sup> (\u221a3x) + Cos-1<\/sup>x = \\(\\frac{\\pi}{2}\\)
\nAnswer:
\nLet cos-1<\/sup> (\u221a3x) = \u03b1 \u21d2 cos \u03b1 = \u221a3x
\nand cos-1<\/sup>x = \u03b2 \u21d2 cos \u03b2 = x
\ncos (\u03b1 + \u03b2) = \\(\\frac{\\pi}{2}\\)
\n\u21d2 \u221a3x . x = \\(\\sqrt{1-3 x^2} \\sqrt{1-x^2}\\) = 0
\n\u21d2 \u221a3x2<\/sup> = \\(\\sqrt{1-3 x^2} \\sqrt{1-x^2}\\)
\n\u21d2 3x4<\/sup> = (1 – 3x2<\/sup>) (1 – x2<\/sup>)
\n= 3x4<\/sup> – 4x2<\/sup> + 1
\n\u21d2 4x2<\/sup> = 1 \u21d2 x = \u00b1\\(\\frac{1}{2}\\) \u21d2 x = – \\(\\frac{1}{2}\\) is not admissible.
\n\u2234 x = \\(\\frac{1}{2}\\)<\/p>\n

(v) sin[Sin-1<\/sup>\\(\\left(\\frac{1}{5}\\right)\\) + Cos-1<\/sup>x] = 1
\nAnswer:
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Intermediate Maths 1A Solutions Chapter 8 Inverse Trigonometric Functions Ex 8(a) to find a better approach to solving the problems. TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) I. 1. Evaluate the following: (i) Sin-1 Answer: (ii) Cos-1 Answer: cos-1 = cos-1 (cos) = (\u2235 cos-1 (cos … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7438"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=7438"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7438\/revisions"}],"predecessor-version":[{"id":7513,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7438\/revisions\/7513"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=7438"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=7438"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=7438"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}