SN Dey Class 11 Solutions Compound Angles Short Answer Type Questions

SN Dey Class 11 Chapter  Solutions. Short answers on compound angles. Compound Angles SATQ. Solutions of Trigonometric Ratios of Compound Angles Short Answer Type Questions.

Ex 1(i):  Prove that sin2α + sin2(120°-α) + sin2(120°+α) = $\frac{3}{2}$

Solution:

 

Ex 1(ii):  Prove that sin2($\frac{\pi}{8}+\frac{\theta}{2}$) – sin2($\frac{\pi}{8}-\frac{\theta}{2}$) = $\frac{1}{\sqrt{2}}$ sinθ

Solution:

sn dey class 11 compound angles short q1.(i) solution

 

Ex 1(iii):  Prove that cos2A + cos2(A+$\frac{\pi}{3}$) + cos2(A-$\frac{\pi}{3}$) = $\frac{3}{2}$ 

Solution:

 

Ex 1(iv):  Prove that tan70° = 2 tan50° + tan20°

Solution:

sn dey class 11 compound angles short q1.(iv) solution

 

Ex 2(i):  Simplify cosA sin(B-C) + cosB sin(C-A) + cosC sin(A-B)

Solution:

cosA sin(B-C) + cosB sin(C-A) + cosC sin(A-B)

= cosA (sinB cosC-cosB sinC) + cosB(sinC cosA – cosC sinA) + cosC(sinA cosB – cosA sinB)

= cosA sinB cosC – cosA cosB sinC + cosA cosB sinC – sinA cosB cosC + sinA cosB cosC – cosA sinB cosC

= 0

 

Ex 2(ii):  Simplify $1+\frac{\sin(A-B)}{\cos A\cos B}$ $+\frac{\sin(B-C)}{\cos B\cos C}$ $+\frac{\sin(C-A)}{\cos C\cos A}$

Solution:

 

Ex 2(iii):  Simplify tan($\frac{\pi}{4}+θ$) + tan($\frac{3\pi}{4}+θ$)

Solution:

 

Ex 2(iv):  Simplify sin(B+C) sin(BC) + sin(C+A) sin(CA) + sin(A+B) sin(AB)

Solution:

sin(B+C) sin(B-C) + sin(C+A) sin(C-A) + sin(A+B) sin(A-B)

= (sin2B – sin2C) + (sin2C – sin2A) + (sin2A – sin2B)

[∵ sin) sin(α-β) = sin2α – sin2β]

= 0