__Multiple Choice Questions__

__Please note that I have replaced theta with A since it is hard to type 5 letters atleast thrice in every sum for one angle and would tend to confuse the reader as well.__

**This is perhaps one of the longest posts I have ever written on the blog!This will consist of hints or solutions to a total of 48 sums!I hope this helps WBCHSE Board students/students preparing for AIJEE who search for solutions of sums online but are disappointed most of the times due to the unavailability or messy solving of sums that you will often find are done in a way that the solver himself might get confused the next time he reads it.**

**It would be a definite pleasure if this helps those students who are in need of solutions to such problems.**

**1. 1 - cos2A = 1 - ( 1 - 2sin^2A) = 2sin^A**

**Answer : Option A**

**2. 1 + cos2A/1 - 2cosA = 1 + ( 1 - 2 sin^2A) / 1 - (1 - 2sin^A)**

**= 2 (1 - sin^2A)/2sin^2A**

**= 2cos^2A/2sin^2A = cot^2A**

**Answer : Option D**

**3. 1 - cos2A/sin 2A = 1 - (1 - 2sin^A)/2sinAcosA = 2sin^A/2sinAcosA = sinA/cosA = tanA**

**Answer : Option C**

**4. 1 + tan^2A/1 -tan^2A = (1 + sin^2A/cos^2A) /(1 - sin^2A/cos^2A) =**

**(cos^2A + sin^2A/cos^2A) / (cos^2A - sin^2A/cos^2A) = 1/cos^2A - sin^2A = 1/cos^2A**

**= sec^A**

**Answer : Option C**

**5. cot 2A + tanA = cos2A/sin2A + sinA/cosA**

**= (1 - 2sin^A/2sinAcosA) + (sinA/cosA)**

**= cosA/2sinAcos^2A**

**= 1/2sinAcosA = cosec2A**

**Answer : Option B**

**6. We know that cos3A = 4cos^3A - 3cosA or 4cos^3A = cos3A +3cosA**

**but it is given in option A that 4cos^3A = 3cos3A + cosA,which is wrong.**

**Therefore,option A is not true.**

**Answer : Option A**

**7. 2cosA = x + 1/x or cosA = (x^2 + 1)/2x or cos^2A = (x^4 + 1 + 2x^2)/4x^2 ....... Eqn 1.**

**Therefore, cos2A =**

**2cos^2A - 1 = 2(x^4 + 1 +2x^2)/4x^2 - 1 [From Eqn 1] = (x^4 + 1)/2x^2 =**

**1/2(x^2 + 1/x^2)**

**Answer : Option B**

**8. Given that sin A = 3/5, cos 2A = 1 - 2sin^2A = 1 - 2 (3/5)^2 = 1 - 18/25 = 7/25**

**Answer : Option D**

**9. Given that tan A = 3, tan 2A = tan (A + A) = (tanA + tan A)/1 - tanA.tanA =**

**(2tanA)/(1 - tan^2A) = (2.3)/(1-9) = -3/4**

**Answer : Option A**

**10. Given that sin A = 3/5,sin 3A = 3sinA - 4sin^3A = 3(3/5) - 4 (3/5)^3 = 9/5 - 108/125**

**= 117/125**

**Answer : Option B**

**Answer : Option C**

**12. Given that tan x = b/a, sin 2x = (2tanx)/(1 + tan^2x) = (2b/a)/(1 + b^2/a^2) = (2ab)/(a^2 +b^2)......Eqn 1**

**Therefore,(a^2 + b^2)sin 2x = (a^2 + b^2).(2ab)/(a^2 + b^2) [From Eqn 1] = 2ab**

**Answer : Option B**

__Very Short Answer Type Questions__**1.Since pi/2 < A < pi, angle A must be lying in the second quadrant. By using formula sin^2 + cos^2 = 1,you will get the value of cos A.Now,since A lies in the second quadrant,cos has to be negative.So,cos = -4/5.So,the answer is -24/25.**

**Similarly,you can solve 2,3 and 4.**

**5. i. R.H.S = 2cot2A = 2cot(A +A) = 2 {(cot^2A - 1)/2cotA} = cotA - 1/cotA = cotA - tanA = L.H.S [Proved]**

**ii. Start with L.H.S,simply cross multiply and use formula of sin2A and cos2A to find your way to the answer.**

**iii.Start with L.H.S,break sin3A and cos3A into their respective formula,you will get**

**(6sinAcosA - 4sin^3AcosA - 4cos^3AsinA)/sinAcosA**

**= {6sinAcosA - 4sinAcosA(sin^2A + cos^2A)}/sinAcosA = 2 = R.H.S**

**iv. Start with L.H.S,break all the given ratios into ratios of sin and cos.You will get**

**(cos^2A - sin^2A/sinAcosA)/(1 - 2sin^2A) = (cos2A)/(sinAcosA.cos2A) = 1/sinAcosA**

**= secAcosecA = R.H.S**

**v. sin^4x + cos^4x = (sin^2x)^2 + (cos^2x)^2 = (sin^2x + cos^x)^2 - 2.sin^2xcos^2x**

**= 1 - 1/2(4sin^2xcos^x) = 1 - 1/2sin^2x = R.H.S**

**vi. Start with R.H.S,simply substitute the value of a with a = sinx + cosx,as given,you will get your way to the answer.**

**vii. tanA(1 + sec2A) = sinA/cosA ( 1 + 1/cos2A) = sinA/cosA {(1 + cos2A)/cos2A}**

**= sinA/cosA { 2 (1 - sin^2A) / (2(1 - sin^2A) - 1} = (2sinAcosA)/(1 - 2sin^2A) = sin2A/cos2A**

**= tan2A = R.H.S**

**viii. Start with L.H.S,rationalise the denominator,you will get (1 + 2sinAcosA)/(cos2A)**

**= 1/cos2A + sin2A/cos2A = sec2A + tan2A = R.H.s**

**6. (96 sin65**

**°**

**sin35**

**°**

**sin80**

**°)**

**/(sin20**

**°**

**+ 2sin80**

**°**

**cos30**

**°)**

**= 96 sin(90**

**° - 25**

**°) sin (90**

**°**

**- 10**

**°**

**) sin 35**

**°/2sin10**

**°cos10**

**° + (**

**√3/2).2sin80**

**°**

**= 96 cos25**

**°cos10**

**°sin35**

**°/2sin10**

**°cos10**

**° +**

**√3sin (90**

**° - 10**

**°)**

**= 48. 2sin35**

**°cos25**

**°cos10**

**°/**

**2sin10**

**°cos10**

**°**

**+**

**√3cos10**

**°**

**= 48. (sin 60**

**° + sin10**

**°)cos10**

**°**

**/ cos10**

**° (2sin10**

**° +**

**√3)**

**= 48 (**

**√3/2 + sin10**

**°)/(2sin10**

**° +**

**√3)**

**= 24**

**(2sin10**

**° +**

**√3)/**

**(2sin10**

**° +**

**√3)**

**= 24**

**Short Answer Type Questions**

**1. i. (Courtesy of this particular sum goes to askiitians forum.)**

**L.H.S = tanA + 2tan2A + 4cot4A = tanA + 2tan2A + 4/tan4A**

**= tanA + 2tan2A + 4/(2tan2A/1 - tan^2A)**

**= tanA + 2tan2A + (4 - 4tan^2A)/2tan2A**

**= tanA + (4tan^2A + 4 - 4tan^2A)/2tan2A**

**= tanA + 2/tan2A = tanA + 2/(2tanA/1 - tan^2A)**

**= tanA + 2/(2tanA/1 - 2tan^2A) = tanA + (1 - tan^2A/tanA) = 1/tanA = cotA = R.H.S**

**ii. This sum has only one step! Using formula cos2A = cos^2A - sin^2A,**

**we have cos^2(A +B) - sin^2(A + B) = cos2Acos2B = R.H.S**

**iii. Start with L.H.S,group 1 with cos,simplify 2A to ratios with only angle A,you will get**

**(2cos^2A + 2sinAcosA)/(2sin^2A + 2sinAcosA) = {2cosA(cosA + sinA)}/{2sinA(cosA + sinA)}**

**= cotA = R.H.S**

**iv. L.H.S = cos^6A + sin^6A = (cos^2A)^3 + (sin^2A)^3**

**= (sin^2A + cos^2A)^3 - 3sin^2Acos^2A(sin^2A + cos^2A)**

**= 1 - 3sin^2Acos^2A......Eqn 1**

**R.H.S = 1/4(1 + 3cos^2 2A) = 1/4 +(3/4).cos^2 2A = 1/4 + (3/4).(cos^2A - sin^2A)^2**

**= 1/4 + 3/4 (cos^4A + sin^4A - 2sin^2Acos^2A)**

**= 1/4 + 3/4{(cos^2A + sin^2A)^2 - 2sin^2Acos^2A - 2sin^2Acos^2A}**

**=1/4 + 3/4(1 - 4sin^2Acoa^2A)**

**= (1 + 3 - 12sin^2Acos^2A)/4 = 4/4(1 - 3**

**sin^2Acos^2A) = 1 -**

**3**

**sin^2Acos^2A......Eqn 2**

**From Eqns 1 and 2,we have L.H.S = R.H.S**

**v. Start with L.H.S,simplify sin2A = 2sinAcosA and cos2A = 1 - 2sin^2A,take common term**

**2sinA.The numerator and denominator will cancel out each other and you will get the answer.**

**vi. Start with L.H.S,simplify and break each ration into ratios of sin and cos,you will get cos2x/sinxcosx = 2cos2x/2sinxcosx = 2cos2x/sin2x = 2cot2x = R.H.S**

**vii. (sec8A - 1)/(sec4A - 1) = (1/cos8A - 1)(1/cos4A - 1) = (1 - cos8A)(cos 4A)/ (cos8A)(1 - cos4A)**

**= {1 - (1 - 2sin^2 4A)}(cos4A)/(cos8A){1 - (1 - 2sin^2A)}**

**= {2sin^2 4A (cos4A)} / (cos8A)(2sin^2 2A)**

**= sin4Acos4A. 2sin2Acos2A/cos8A.sin2A.sin2A**

**= 2sin4Acos4A.cos2A/sin2A.cos8A**

**=2sin4Acos4A/cos8A . 1/(sin2A/cos2A)**

**= sin8A/cos8A .**

**1/(sin2A/cos2A)**

**= tan8A/tan2A = R.H.S**

**viii. cos 4x - cos 4y = (2cos^2 2x -1) - (2cos^2 2y - 1) = 2(cos^2 2x - cos^2 2y)**

**=2(cos2x + cos2y)(cos2x - cos2y)**

**= 2 {(2cos^2x - 1) + ( 1 - 2sin^2y)}{(2cos^2x - 1) - (2cos^2y - 1)}**

**= 2.2(cos^2x - sin^2y) 2.(cos^2x - cos^2y)**

**= 8 (cosx + siny)(cosx - siny)(cosx - cosy)(cosx + cosy)**

**(Courtesy to some random solver on an Yahoo forum)**

**ix. Start with L.H.S,simply break sin2A and cos2A and take a common term at both numerator and denominator such that all the other terms cancel out and you are left with only**

**cosA/sinA = cotA.**

**x. The courtesy of this particular sum goes to Meritnation.**

**(tan5A + tan3A)/(tan5A - tan3A) = (sin5A/cos5A + sin3A/cos3A)/(sin5A/cos5A - sin3A/cos3A)**

**= (sin5Acos3A + sin3Acos5A)/(sin5Acos3A - sin3Acs5A)**

**= sin (5A + 3A)/sin(5A - 3A) = (sin8A)/sin2A = (2sin4Acos4A)/sin2A**

**= (2.2sin2Acos2A.cos4A)/sin2A = 4cos4Acos2A = R.H.S**

**The rest of the solutions will be up here by 14th June.(Please excuse the delay,I have got a big homework on Geometric Progressions.)**

**xi.**

**xii.**

**xiii.**

**xiv.**

Help full solution.. Hope it gonna help lots of students..

ReplyDeletehttp://zigzacmania.blogspot.in/

Thank You Anjali. I really hope it does. :)

Delete