If tan -2021- show that1-sin -cos- 1-sin-cos-37-

Let us consider a right △ABC right angled at B and ∠C=θ. Now, we know that tan θ = nbsp;ABBC = 2021 So, if AB = 20k, then BC = 21k, where k is a positive num ...

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Standard X

Mathematics

Complementary Trigonometric Ratios

If tan θ=2021...

Question

If tan θ = $\frac{20}{21}$ , show that $\frac{(1 - sinθ + cosθ)}{(1 + sinθ + cosθ)} = \frac{3}{7} .$

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Solution

Let us consider a right $△$ ABC right angled at B and $∠ C = θ$ .
Now, we know that tan $θ$ = $\frac{A B}{B C}$ = $\frac{20}{21}$

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC²= (20k)²+ (21k)²
⇒ AC² = 841k²
⇒ AC = 29k
Now, sin $θ$ = $\frac{A B}{A C} = \frac{20}{29}$ and cos $θ$ = $\frac{B C}{A C} = \frac{21}{29}$

Substituting these values in the given expression, we get:
$LHS = \frac{1 - \sin θ + \cos θ}{1 + \sin θ + \cos θ} = \frac{1 - \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29 - 20 + 21}{29}}{\frac{29 + 20 + 21}{29}} = \frac{30}{70} = \frac{3}{7} = RHS$
∴ LHS = RHS

Hence proved.

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Similar questions

Q. If

\tan θ = \frac{20}{21}

, show that

\frac{1 - \sin θ + \cos θ}{1 + \sin θ + \cos θ} = \frac{3}{7}

Q. Prove each of the following identities:

(i) \sqrt{\frac{1 + \sin θ}{1 - \sin θ}} = (\sec θ + \tan θ) (ii) \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = (cosec θ - \cot θ) (iii) \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ

Q. If

cot θ = \frac{7}{8}

, evaluate:

\frac{(1 + sin θ) (1 - sin θ)}{(1 + cos θ) (1 - cos θ)}

Q. Prove the following trigonometric identities.

(i)

\sqrt{\frac{\sec θ - 1}{\sec θ + 1}} + \sqrt{\frac{\sec θ + 1}{\sec θ - 1}} = 2 cosec θ

(ii)

\sqrt{\frac{1 + \sin θ}{1 - \sin θ}} + \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} = 2 \sec θ

(iii)

\sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ

(iv)

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

(v)

\frac{\sin θ + 1 - \cos θ}{\cos θ - 1 + \sin θ} = \frac{1 + \sin θ}{\cos θ}

\frac{\sin θ}{1 - \cot θ} + \frac{\cos θ}{1 - \tan θ}

is equal to

(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ − cos θ

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If tan θ = 2021, show that1-sinθ+cosθ1+sinθ+cosθ=37.

If tan θ = $\frac{20}{21}$ , show that $\frac{(1 - sinθ + cosθ)}{(1 + sinθ + cosθ)} = \frac{3}{7} .$