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Lesson Example Discussion Quiz: Class Homework

Step-2

Title: Trigonometry Identities (quotient , co-function)

Grade: 10-a Lesson: S3-L4

Explanation: Hello Students, time to practice and review the steps for the problem.

Lesson Steps

Step	Type	Explanation	Answer
1	Problem	Find the value of \$"cosA" times 1/("sinB") - "cosA" times 1/ ("cosecB")\$ such that A and B are two complementary angles.
2	Step	The given expression	\$"cosA" times 1/("sinB") - "cosA" times 1/ ("cosecB")\$
3	Hint	Since A and B are complementary angles, we have \$B = 90^∘ − A\$. Substituting this into the expression, we get:	\$"cosA" times 1/("sin"(90^∘ - A)) − "cosA" times 1/("csc"(90^∘ - A))\$
4	Step	Now, let’s simplify this expression:	1.\$sin(90^∘ − A) = cos(A)\$ (using the complementary angle identity). 2. \$csc(90^∘−A)= 1/sin(90^∘−A) = 1/cosA\$ (using the reciprocal identity)
5	Step	Let’s substitute these relationships into the given expression	⇒ \$"cosA" times 1/("cosA") − "cosA" times 1/(1/("cosA"))\$ ⇒ \$1 − "cosA" times "cosA"\$ ⇒ \$1 - "cos"^2("A")\$
6	Hint	Since \$cos^2(A) + sin^2(A) = 1 \$(using the Pythagorean identity), we can rewrite \$cos^2(A)\$ as \$1 - sin^2(A) \$. Substituting this into the expression, we get:	\$1 − (1 − "sin"^2("A"))\$ ⇒ \$1 - 1 + "sin"^2("A")\$ ⇒ \$"sin"^2("A")\$
7	Step	The value of the expression for complementary angles A and B is \$"sin"^2("A")\$.
8	Choice.A	Wrong: The expression simplifies to \$"sin"^2("A")\$ rather than \$"tan"^2("A")\$	\$"tan"^2("A")\$
9	Choice.B	The calculation is correct, which is why this is also correct	\$"sin"^2("A")\$
10	Choice.C	Our derivation simplifies Option C from cotangent squared to sine squared. However, since cotangent is not equal to sine, Option C is incorrect	\$"cot"^2("A")\$
11	Choice.D	\$"cos"^2("A")\$ is incorrect because the expression ultimately simplifies to \$"sin"^2("A") − "cos"^2("A")\$ rather than \$"cos"^2("A")\$	\$"cos"^2("A")\$
12	Answer	Option	B
13	Sumup	Can you summarize what you’ve understood in the above steps?

Step

Type

Explanation

Answer

Problem

Find the value of \$"cosA" times 1/("sinB") - "cosA" times 1/ ("cosecB")\$ such that A and B are two complementary angles.

Step

The given expression

\$"cosA" times 1/("sinB") - "cosA" times 1/ ("cosecB")\$

Hint

Since A and B are complementary angles, we have \$B = 90^∘ − A\$. Substituting this into the expression, we get:

\$"cosA" times 1/("sin"(90^∘ - A)) − "cosA" times 1/("csc"(90^∘ - A))\$

Step

Now, let’s simplify this expression:

1.\$sin(90^∘ − A) = cos(A)\$
(using the complementary angle identity).
2. \$csc(90^∘−A)= 1/sin(90^∘−A) = 1/cosA\$ (using the reciprocal identity)

Step

Let’s substitute these relationships into the given expression

⇒ \$"cosA" times 1/("cosA") − "cosA" times 1/(1/("cosA"))\$
⇒ \$1 − "cosA" times "cosA"\$
⇒ \$1 - "cos"^2("A")\$

Hint

Since \$cos^2(A) + sin^2(A) = 1 \$(using the Pythagorean identity), we can rewrite \$cos^2(A)\$ as \$1 - sin^2(A) \$. Substituting this into the expression, we get:

\$1 − (1 − "sin"^2("A"))\$

⇒ \$1 - 1 + "sin"^2("A")\$

⇒ \$"sin"^2("A")\$

Step

The value of the expression for complementary angles A and B is \$"sin"^2("A")\$.

Choice.A

Wrong: The expression simplifies to \$"sin"^2("A")\$ rather than \$"tan"^2("A")\$

\$"tan"^2("A")\$

Choice.B

The calculation is correct, which is why this is also correct

\$"sin"^2("A")\$

Choice.C

Our derivation simplifies Option C from cotangent squared to sine squared. However, since cotangent is not equal to sine, Option C is incorrect

\$"cot"^2("A")\$

Choice.D

\$"cos"^2("A")\$ is incorrect because the expression ultimately simplifies to \$"sin"^2("A") − "cos"^2("A")\$ rather than \$"cos"^2("A")\$

\$"cos"^2("A")\$

Answer

Option

Sumup

Can you summarize what you’ve understood in the above steps?