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:

 

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:

 

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(B−C) + sin(C+A) sin(C−A) + sin(A+B) sin(A−B)

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(α+β) sin(α-β) = sin²α – sin²β]

= 0