SN Dey Class 11 Compound Angles Long Answer Type Questions Solutions

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Ex 1:  Prove that $\tan(A+B)-\tan(A-B)=\frac{\sin 2B}{\cos^2B-\sin^2A}$

Solution:

sn dey class 11 compound angles long answer q1 solution

 

Ex 2:  If $\tan \beta = \frac{n\sin\alpha \cos\alpha}{1-n\sin^2\alpha},$ Show that, tan(α-β) = (1-n) tan α.

Solution:

sn dey class 11 compound angles Q2 long answer solution

 

Ex 3:  If m tan(θ-30°) = n tan(θ+120°), show that, 2 cos2θ = $\frac{m+n}{m-n}$.

Solution:

 

Ex 4:  If cos(β-γ) + cos(γ-α) + cos(α-β) = $-\frac{3}{2}$, show that, cosα + cosβ + cosγ=0 and sinα + sinβ + sinγ=0. Hence deduce that

cos(β-γ) = cos(γ-α) = cos(α-β) = $-\frac{1}{2}$.

Solution:

 

Ex 5:  If α ≠ β and a tanα + b tanβ = (a+b) tan $\frac{\alpha+\beta}{2}$, show that, $\frac{\cos \alpha}{\cos \beta}=\frac{a}{b}$.

Solution:

sn dey class 11 compound angles long answer q5 solution

 

Ex 6:  If sinθ=k sin(θ+φ), show that, tan(θ+φ)=$\frac{\sin \phi}{\cos \phi -k}$.

Solution:

 

Ex 7:  If $\frac{\cot(\alpha-\beta)}{\cot \alpha}$+$\frac{\cos^2 \gamma}{\cos^2 \alpha}=1$, show that,  tan2γ + tanα cotβ =0.

 Solution:

 

Ex 8:  If tanθ=$\frac{x \sin \alpha+y\sin \beta}{x \cos \alpha+y\cos \beta}$, prove that, x sin(θ-α) + y sin(θ-β) =0.

Solution: