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2 years ago

Tan^2 70° -sec^2 70°

Mathematics

1 answer:

kipiarov [429]2 years ago

4 0

Answer:

-1

Step-by-step explanation:

We need to find the value of tan² 70° -sec² 70°.

We know that,

$\sec^2\theta-\tan^2\theta=1$ ...(1)

We can write the given expression as :

tan² 70° -sec² 70° = -(-tan² 70° +sec² 70°)

=-(sec² 70°-tan² 70°)

Using identity (1)

=-(1)

= -1

Hence, the value of the given expression is -1.

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Simplify, state all restrictions.

Kipish [7]

The simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$

<h3>How to simplify the expression?</h3>

The expression is given as:

$\frac{x - y}{4x^2 - 8xy + 3y^2} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{x^2 - y^2} -1$

Express x^2 - y^2 as (x + y)(x - y) and factorize other expressions

$\frac{x - y}{(2x - y)(2x - 3y)} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Rewrite the expression as products

$\frac{x - y}{(2x - y)(2x - 3y)} \times \frac{2x - 3y}{2x + y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Cancel out the common factors

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{4x^2 - y^2}{(x + y)} -1$

Express 4x^2 - y^2 as (2x - y)(2x + y)

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{(2x - y)(2x + y)}{(x + y)} -1$

Cancel out the common factors

$\frac{1}{x + y} -1$

Take the LCM

$\frac{1 - x - y}{x + y}$

Hence, the simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$

Read more about expressions at:

brainly.com/question/723406

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Tan^2 70° -sec^2 70°​

Tan^2 70° -sec^2 70°