If sin 2 A -sin 2 B- provethat - tan A - B -tan A - B -1-1

We have sin2A=λ sin2B ⇒λ =sin2A/sin2B Now, λ +1/λ 1=sin2A/sin2B+1/sin2A/sin2B 1 =sin2A+sin2B/sin2B/sin2A sin2B/sin2B =sin2A+sin2B/sin2A sin2B =2sin ( 2A+2B/ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If sin 2 A =λ...

Question

If $s i n 2 A = λ s i n 2 B, p r o v e t h a t : \frac{t a n (A + B)}{t a n (A - B)} = \frac{λ + 1}{λ - 1}$

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Solution

We have
$s i n 2 A = λ s i n 2 B \Rightarrow λ = \frac{s i n 2 A}{s i n 2 B}$
Now,
$\frac{λ + 1}{λ - 1} = \frac{\frac{s i n 2 A}{s i n 2 B} + 1}{\frac{s i n 2 A}{s i n 2 B} - 1} = \frac{\frac{s i n 2 A + s i n 2 B}{s i n 2 B}}{\frac{s i n 2 A - s i n 2 B}{s i n 2 B}} = \frac{s i n 2 A + s i n 2 B}{s i n 2 A - s i n 2 B} = \frac{2 s i n (\frac{2 A + 2 B}{2}) c o s (\frac{2 A - 2 B}{2})}{2 s i n (\frac{2 A - 2 B}{2}) c o s (\frac{2 A + 2 B}{2})} = \frac{s i n (A + B) c o s (A - B)}{s i n (A - B) c o s (A + B)} ∴ \frac{λ + 1}{λ - 1} = \frac{t a n (A + B)}{t a n (A - B)} \Rightarrow \frac{t a n (A + B)}{t a n (A - B)} = \frac{λ + 1}{λ - 1}$

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Q. If

sin 2 A = λ sin 2 B

, then write the value of

\frac{λ + 1}{λ - 1}

Q. If sin 2A = λ sin 2B, then write the value of

\frac{λ + 1}{λ - 1}

If \sin 2 A = λ \sin 2 B, prove that \frac{\tan (A + B)}{\tan (A - B)} = \frac{λ + 1}{λ - 1}

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If sin2A=λ sin2B,provethat:tan(A+B)tan(A−B)=λ+1λ−1

We have sin2A=λ sin2B⇒λ=sin2Asin2B Now, λ+1λ−1=sin2Asin2B+1sin2Asin2B−1=sin2A+sin2Bsin2Bsin2A−sin2Bsin2B=sin2A+sin2Bsin2A−sin2B=2sin(2A+2B2)cos(2A−2B2)2sin(2A−2B2)cos(2A+2B2)=sin(A+B)cos(A−B)sin(A−B)cos(A+B)∴ λ+1λ−1=tan(A+B)tan(A−B)⇒tan(A+B)tan(A−B)=λ+1λ−1

If $s i n 2 A = λ s i n 2 B, p r o v e t h a t : \frac{t a n (A + B)}{t a n (A - B)} = \frac{λ + 1}{λ - 1}$