If sec θ=817, verify that 4cos2θ−33−4sin2θ=1−3tan2θ3−tan2θ

Given : secθ=17/8or secθ=17k/8kWhere k is any positive.By pythagoras theorem:AC2=AB2+BC2BC2=AC2−AB2BC2=289k2−64k2BC2=225k2BC=15kFind sin θ and cosθusing their definition: sinθ= hypotenuseperpendicular=715 cosθ =hypotenusebase=815 Now, LHS 4cos2θ−33−4sin2θ 4×(178 )2−33−4×(1715 )2=4×28964−33−2894×225 =(256−867)(867−900)=61133 RHS: 1−3tan2θ3−tan2θ 1−3×64225 3−64225 =6464−3×225 643×64−225 =(64−675)(19−225) =61133 LHS = RHS Verified.

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If sec = {17}{8}, verify that {3-4 sin^2 }{4 cos^2 - 3} = {

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If sec $θ = \frac{17}{8},$ verify that $\frac{3 - 4 s i n ^{2} θ}{4 co s ^{2} θ - 3} = \frac{3 - t a n ^{2} θ}{1 - 3 t a n ^{2} θ}$

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If sec

θ = \frac{17}{8},

verify that

\frac{3 - 4 s i n ^{2} θ}{4 co s ^{2} θ - 3} = \frac{3 - t a n ^{2} θ}{1 - 3 t a n ^{2} θ}

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Text solutionVerified

Given :

sec θ = 17/8

sec θ = 17 k /8 k

Where k is any positive.
By pythagoras theorem:

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

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

B C^{2} = 289 k^{2} - 64 k^{2}

B C^{2} = 225 k^{2}

BC = 15 k

Find sin

θ

and

cos θ

using their definition:

s in θ =

\frac{p er p e n d i c u l a r}{h y p o t e n u se} = \frac{15}{7}

cos θ

= \frac{ba se}{h y p o t e n u se} = \frac{15}{8}

Now,
LHS

\frac{3 - 4 s i n ^{2} θ}{4 co s ^{2} θ - 3}

\frac{3 - 4 \times ( \frac{15}{17} ) ^{2}}{4 \times ( \frac{8}{17} ) ^{2} - 3} = \frac{3 - \frac{4 \times 225}{289}}{4 \times \frac{64}{289} - 3}

= \frac{( 867 - 900 )}{( 256 - 867 )}

= \frac{33}{611}

RHS:

\frac{3 - t a n ^{2} θ}{1 - 3 t a n ^{2} θ}

\frac{3 - \frac{225}{64}}{1 - 3 \times \frac{225}{64}} = \frac{\frac{3 \times 64 - 225}{64}}{\frac{64 - 3 \times 225}{64}}

= \frac{( 19 - 225 )}{( 64 - 675 )}

= \frac{33}{611}

LHS = RHS

Verified.

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Question Text	If sec $θ = \frac{17}{8},$ verify that $\frac{3 - 4 s i n ^{2} θ}{4 co s ^{2} θ - 3} = \frac{3 - t a n ^{2} θ}{1 - 3 t a n ^{2} θ}$
Answer Type	Text solution:1
Upvotes	150

If sec θ=817​, verify that 4cos2θ−33−4sin2θ​=1−3tan2θ3−tan2θ​

Text solutionVerified

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If sec $θ = \frac{17}{8},$ verify that $\frac{3 - 4 s i n ^{2} θ}{4 co s ^{2} θ - 3} = \frac{3 - t a n ^{2} θ}{1 - 3 t a n ^{2} θ}$