Question

Evaluate : (i) sin (90° – A) cos A + cos (90° – A) sin A
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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Trigonometric Identities has been used. We see that we have to evaluate the given expression. We can do it by simplifying the equation part by part. We can use Trigonometric identities to simplify this and again use it to get the final answer.

Let's do it !!

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★ Identities used :-

$\\\;\boxed{\sf{\pink{\sin\:\theta\;=\;\bf{\cos\:(90^{\circ}\;-\;\theta)}}}}$

$\\\;\boxed{\sf{\pink{\sin^{2}\:\theta\;+\;\cos^{2}\:\theta\;=\;\bf{1}}}}$

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★ Solution :-

Given expression,

→ sin(90° - A) cos A + cos(90° - A) sin A

We already know the identity that,

$\\\;\sf{:\rightarrow\;\;\sin\:\theta\;=\;\bf{\orange{\cos\:(90^{\circ}\;-\;\theta)}}}$

From this we can derive that,

$\\\;\sf{:\rightarrow\;\;\cos\:\theta\;=\;\bf{\orange{\sin\:(90^{\circ}\;-\;\theta)}}}$

From given expression,

$\\\;\sf{\Longrightarrow\;\;\sin\:(90^{\circ}\;-\;A)\;\cos\:A\;+\;\cos\:(90^{\circ}\;-\;A)\;\sin\:A}$

By applying the value of cos A, we get

$\\\;\sf{\Longrightarrow\;\;\cos\:A\;\cos\:A\;+\;\cos\:(90^{\circ}\;-\;A)\;\sin\:A}$

$\\\;\sf{\Longrightarrow\;\;\cos\:A\:\times\:\cos\:A\;+\;\cos\:(90^{\circ}\;-\;A)\;\sin\:A}$

$\\\;\sf{\Longrightarrow\;\;\cos^{2}\:A\;+\;\cos\:(90^{\circ}\;-\;A)\;\sin\:A}$

Now using the value of sin A, we get

$\\\;\sf{\Longrightarrow\;\;\cos^{2}\:A\;+\;\sin\:A\;\sin\:A}$

$\\\;\sf{\Longrightarrow\;\;\cos^{2}\:A\;+\;\sin\:A\;\times\;\sin\:A}$

$\\\;\sf{\Longrightarrow\;\;\cos^{2}\:A\;+\;\sin^{2}\:A}$

We already know that,

$\\\;\sf{\orange{\sin^{2}\:\theta\;+\;\cos^{2}\:\theta\;=\;\bf{1}}}$

Now applying this, we get

$\\\;\sf{\Longrightarrow\;\;\cos^{2}\:A\;+\;\sin^{2}\:A}$

$\\\;\sf{\Longrightarrow\;\;\sin^{2}\:A\;+\;\cos^{2}\:A}$

$\\\;\bf{\Longrightarrow\;\;\red{1}}$

Thus, 1 is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\:Answer\;=\;\bf{\purple{1}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;\sec^{2}\:\theta\;=\;1\;+\;\tan^{2}\:\theta}$

$\\\;\sf{\leadsto\;\;cosec^{2}\:\theta\;=\;1\;+\;\cot^{2}\:\theta}$

$\\\;\sf{\leadsto\;\;cosec\:\theta\;=\;\sec(90^{\circ}\;-\;\theta)}$

$\\\;\sf{\leadsto\;\;\tan\:\theta\;=\;\cot(90^{\circ}\;-\;\theta)}$

Answer 2

Answer:

see this image ⬆⬆⬆⬆⬆⬆

Step-by-step explanation:

hope it's help you