Question

answer 3rd question in that​

Answer 1

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Given: The term √1-sin²100 sec 100°

To find: The value of the given term?

Solution:

• Now the term provided istrigonometric, which is :

√(1-sin²100°) ×sec 100°

• Now we known sin^2x+cos^2x=1
• So applying it, we get:

√(1-sin²100°) x sec 100°= √(cos²100°) x sec 100°

=cos

100° x sec 100°

• Now we know that cos X × sec x=1

Cos 100°x sec 100°=1

Answer:

So the value of √1-sin²100 sec 100° is 1.

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Answer 2

(3.)

TO FIND :–

• Value of $\bf \sqrt{1 - { \sin}^{2}({100}^{ \circ})} . \sec( {100}^{ \circ} ) = ?$

SOLUTION :–

• Let –

$\\ \bf \implies y = \sqrt{1 - { \sin}^{2}({100}^{ \circ})} . \sec( {100}^{ \circ} )$

• Using identity –

$\\ \bf \implies \cos( \theta) = \sqrt{1 - { \sin}^{2}( \theta)}$

$\\ \bf \implies y ={ \cos}({100}^{ \circ}). \sec( {100}^{ \circ} )$

• Using identity –

$\\ \bf \implies \sec( \theta) = \dfrac{1}{ \cos( \theta) }$

$\\ \bf \implies y ={ \cos}({100}^{ \circ}). \dfrac{1}{ \cos( {100}^{ \circ}) }$

$\\ \large \implies{ \boxed{ \bf y =1}}$

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(4.)

GIVEN :–

$\\ \bf \implies \sec( \theta) + \tan( \theta) =P$

TO FIND :–

$\\ \bf \implies \sec( \theta) - \tan( \theta) =?$

SOLUTION :–

• We know that –

$\\ \bf \implies \sec^{2} ( \theta) - \tan^{2} ( \theta) =1$

• Using identity –

$\\ \bf \implies a^{2} - b^{2} =(a + b)(a - b)$

$\\ \bf \implies \{\sec( \theta) + \tan( \theta) \} \{ \sec( \theta) - \tan( \theta)\} =1$

$\\ \bf \implies P\{ \sec( \theta) - \tan( \theta)\} =1$

$\\ \large\implies{ \boxed{ \bf \sec( \theta) - \tan( \theta) = \dfrac{1}{P}}}$