If 3cotθ=4, show that (1+tan2θ)(1−tan2θ)=(cos2θ−sin2θ).

Given: 3cotθ=4 or cotθ=3k4kWhere k is any positive.By pythagoras theorem ;AC2=AB2+BC2 =16k2+9k2AC=5k Find sinθ and cosθ using their definition cosθ=hypotenuseperpendicular =54 sinθ=hypotenusebase =53 Now, LHS = (1+tan2θ)(1−tan2θ) =1+(43)21−(43)2 =1+1691−169=167×2516 =257 Again, RHS = (cos2θ−sin2θ) = (54 )2−(53 )2 =2516−259 =257 Therefore, RHS = LHS Hence proved.

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If 3 = 4 , show that {(1-^2 )}{(1+^2 )}=(^2 - ^2 ) .

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If $3 cot θ = 4$ , show that $\frac{( 1 - tan ^{2} θ )}{( 1 + tan ^{2} θ )} = (cos^{2} θ - sin^{2} θ)$ .

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3 cot θ = 4

, show that

\frac{( 1 - tan ^{2} θ )}{( 1 + tan ^{2} θ )} = (cos^{2} θ - sin^{2} θ)

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Given:

3 cot θ = 4

cot θ = \frac{4 k}{3 k}

Where

k

is any positive.

By pythagoras theorem ;

A C^{2} = A B^{2} + B C^{2}

= 16 k^{2} + 9 k^{2}

A C = 5 k

Find

sin θ

and

cos θ

using their definition

cos θ = \frac{p er p e n d i c u l a r}{h y p o t e n u se} = \frac{4}{5}

sin θ = \frac{ba se}{h y p o t e n u se} = \frac{3}{5}

Now,
LHS =

\frac{( 1 - t a n ^{2} θ )}{( 1 + t a n ^{2} θ )}

= \frac{1 - ( \frac{3}{4} ) ^{2}}{1 + ( \frac{3}{4} ) ^{2}}

= \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{7}{16} \times \frac{16}{25}

= \frac{7}{25}

Again,
RHS =

(co s^{2} θ - s i n^{2} θ)

= (\frac{4}{5})^{2} - (\frac{3}{5})^{2}

= \frac{16}{25} - \frac{9}{25}

= \frac{7}{25}

Therefore, RHS = LHS
Hence proved.

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Question Text	If $3 cot θ = 4$ , show that $\frac{( 1 - tan ^{2} θ )}{( 1 + tan ^{2} θ )} = (cos^{2} θ - sin^{2} θ)$ .
Answer Type	Text solution:1
Upvotes	150

If 3cotθ=4, show that (1+tan2θ)(1−tan2θ)​=(cos2θ−sin2θ).

Text solutionVerified

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If $3 cot θ = 4$ , show that $\frac{( 1 - tan ^{2} θ )}{( 1 + tan ^{2} θ )} = (cos^{2} θ - sin^{2} θ)$ .