Question

-1 cm है। sinQ और cosQ कमान ज्ञात कीजिए।
एक वृत्त की परिधि ज्ञात कीजिए जिसका क्षेत्रफल 6.16 cm है।
मक क्या होगा
?​

Answer 1

(B) Surya Public School, Ahmedabad requires two sports coaches, one

(B) Surya Public School, Ahmedabad requires two sports coaches, onemale and one female. Each should be a degree holder in physical

(B) Surya Public School, Ahmedabad requires two sports coaches, onemale and one female. Each should be a degree holder in physicaleducation with at least 2 years of coaching experience with children. You

(B) Surya Public School, Ahmedabad requires two sports coaches, onemale and one female. Each should be a degree holder in physicaleducation with at least 2 years of coaching experience with children. Youhave seen their advertisement and you know you have these qualifications.

(B) Surya Public School, Ahmedabad requires two sports coaches, onemale and one female. Each should be a degree holder in physicaleducation with at least 2 years of coaching experience with children. Youhave seen their advertisement and you know you have these qualifications.Write an application in 120-150 words along with your resume. You are

(B) Surya Public School, Ahmedabad requires two sports coaches, onemale and one female. Each should be a degree holder in physicaleducation with at least 2 years of coaching experience with children. Youhave seen their advertisement and you know you have these qualifications.Write an application in 120-150 words along with your resume. You areSwarna/ Swayam, 2/264 MG Road, Ahmedabad.

Answer 2

Answer:

sinθ+cosθ=a

secθ+cscθ=b

\sf\underline \red{ To\:Find}

ToFind

We have to find the value of b(a²-1)

\sf\underline \pink{ Solution }

Solution

By putting the given values

\begin{gathered}:\implies\sf\ \ b(a^2-1)\\ \\ \\ :\implies\sf\ \ sec\theta+csc\theta\big\{(sin\theta+cos\theta)^2-1)\big\}\\ \\ \\ \bullet\sf\ sec\theta=\dfrac{1}{cos\theta}\ \ ;\ csc\theta=\dfrac{1}{sin\theta}\\ \\ \\ :\implies\sf\ \dfrac{1}{cos\theta}+\dfrac{1}{sin\theta}\big\{sin^2\theta+cos^2\theta+2sin\theta cos\theta-1\big\}\\ \\ \\ \bullet\sf\ \ sin^2\theta+cos^2\theta=1\\ \\ \\ :\implies\sf\dfrac{sin\theta+cos\theta}{sin\theta\ cos\theta}\big\{\cancel{1}+2sin\theta\ cos\theta \cancel{-1}\big\}\\ \\ \\ :\implies\sf\ \dfrac{sin\theta+cos\theta}{\cancel{sin\theta cos\theta}}\times 2\cancel{sin\theta cos\theta}\\ \\ \\ :\implies\sf\ \ 2(sin\theta+cos\theta)\end{gathered}

:⟹ b(a

2

−1)

:⟹ secθ+cscθ{(sinθ+cosθ)

2

−1)}

∙ secθ=

cosθ

1

; cscθ=

sinθ

1

:⟹

cosθ

1

+

sinθ

1

{sin

2

θ+cos

2

θ+2sinθcosθ−1}

∙ sin

2

θ+cos

2

θ=1

:⟹

sinθ cosθ

sinθ+cosθ

{

1

+2sinθ cosθ

−1

}

:⟹

sinθcosθ

sinθ+cosθ

×2

sinθcosθ

:⟹ 2(sinθ+cosθ)

\underline{\bigstar{\blue{\sf\ \ b(a^2-1)= 2(sin\theta+cos\theta)}}}

★ b(a

2

−1)=2(sinθ+cosθ