Plus One Math's Solution Miscellaneous Chapter 3 Trigonometric Functions

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The solutions for the
 Trigonometric Functions are not readily available. While many schools may have a ready-made solution for this set in their school textbook, some might not. This is where our solution will be useful. In this article, you will find detailed solutions provided by us for the above set.Trigonometric Functions (Key Concept Reference) describes some basic and advanced uses of trigonometric functions, including identities, graph transformations, inverse functions, solutions of triangles, and polar coordinates.

Ncert Plus one Maths chapter-wise textbook solution for chapter 3 Trigonometric Functions Exercise 3.1. It contains detailed solutions for each question which have prepared by expert teachers to make each answer easily understand the students. they are well arranged solutions so that students would be able to understand easily.

BoardSCERT, Kerala
Text BookNCERT Based
ClassPlus One 
SubjectMath's Textbook Solution
ChapterChapter 3
ExerciseMiscellaneous
Chapter NameTrigonometric Functions
CategoryPlus One Kerala


Kerala Syllabus Plus One Math's Textbook Solution Chapter  3 Trigonometric Functions Miscellaneous Exercises 


Chapter  3 Trigonometric Functions Textbook Solution



Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  3 Trigonometric Functions Miscellaneous Exercise

Prove that: 

L.H.S.

= 0 = R.H.S


Prove that: (sin 3+ sin x) sin + (cos 3– cos x) cos = 0

L.H.S.

= (sin 3+ sin x) sin + (cos 3– cos x) cos x

= RH.S.


Prove that: 

L.H.S. = 

Prove that: 

L.H.S. = 


Prove that: 

It is known that.

∴L.H.S. = 


Prove that: 

numerator=(sin7x+sin5x)+(sin9x+sin3x)

=2sin(12x/2)cos(2x/2)+2sin(12x/2)cos(6x/2)              (sinA+sinB=2sin(A+B)/2 cos(A-B)/2 )

=2sin6x cosx+2sin6x cos3x

=2sin6x(cosx+cos3x)

=2sin6x(2cos(4x/2)cos(2x/2))                           (cosA+cosB=2cos(A+B)/2 cos(A-B)/2)

=4sin6x cos2x cosx

denominator=(cos7x + cos5x)+(cos9x+cos3x)          (cosA+cosB=2cos(A+B)/2 cos(A-B)/2)

=2cos(12x/2)cos(2x/2)+2cOs(12X/2)Cos(6X/2)

=2cos6x cosx +2cos6x cos3x

=2cos6x(cosx+cos3x)

=2cos6x * 2cos(4x/2)cos(2x/2)

=4cos6x cos2x cosx

LHS=(4sin6x cos2x cosx)/(4cos6x cos2x cosx)

=sin6x/cos6x

=tan6x


Prove that: 

L.H.S. = 


x in quadrant II

Here, x is in quadrant II.

i.e.,

Therefore,  are all positive.

As x is in quadrant II, cosx is negative.

Thus, the respective values of are.


Find for x in quadrant III

 

Here, x is in quadrant III.

Therefore,  and  are negative, whereasis positive.

Now, 

Thus, the respective values of are.


Question 10:

Find for x in quadrant II

Here, x is in quadrant II.

Therefore,, and  are all positive.

 [cosx is negative in quadrant II]

Thus, the respective values of are .



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Chapter 3 Trigonometric Functions Miscellaneous Solution


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