Gender Advocacy in Nepal

Gender advocacy refers to the discussion in various aspect based on gender in society. This note provides further information about gender advocacy in Nepal.

Summary

Gender advocacy refers to the discussion in various aspect based on gender in society. This note provides further information about gender advocacy in Nepal.

Things to Remember

  • Gender advocacy refers to the discussion in various aspect based on gender in society.
  • The depth of subject matter and impact to the society should be made clear by gender advocacy.
  • The gender advocacy becomes highly successful if all resources are collected and mobilized on time.
  • Monitoring and evaluation is necessary in order to identify the impact of the programme under gender advocacy.

MCQs

No MCQs found.

Subjective Questions

Q1:

Multiply:

(sinA + sinB) (sinA - sinB)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>Here, (sinA + sinB) (sinA - sinB)</p> <p>= sinA (sinA - sinB) + sinB (sinA - sinB)</p> <p>= sin<sup>2</sup>A - sinA.sinB + sinA.sinB - sin<sup>2</sup>B</p> <p>= sin<sup>2</sup>A - sin<sup>2</sup>B</p>

Q2:

Multiply:

(1 - cosθ) ( 1 + cosθ)

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>Here, ( 1 - cos&theta;) ( 1 + cos&theta;)</p> <p> = 1 - cos<sup>2</sup>&theta;</p> <p> = sin<sup>2</sup>&theta;. Ans</p>

Q3:

Multiply:

(1 + tanθ) (1 - tanθ) (1 + tan2θ)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>Here, ( 1+ tan&theta;) ( 1 - tan&theta;) ( 1 + tan<sup>2</sup>&theta;)</p> <p>= (1 - tan<sup>2</sup>&theta;) (1 + tan<sup>2</sup>&theta;) = 1 - tan<sup>4</sup>&theta; Ans.</p>

Q4:

Factorize :

cos2A - sin2A


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>cos<sup>2</sup>A - sin<sup>2</sup>A = ( cosA - sinA) (cosA + sinA). Ans.</p>

Q5:

Prove the following:

(1 + tan2A) cos2A = 1


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>L. H. S. = (1 + tan<sup>2</sup>A) cos<sup>2</sup>A</p> <p>= sec<sup>2</sup>A. cos<sup>2</sup>A</p> <p>= \(\frac{1}{cos^2A}\). cos<sup>2</sup>A</p> <p>= 1 = RHS Proved.</p>

Q6:

Prove the following:

\(\frac{1}{cos^2A}\) - \(\frac{1}{cot^2A}\) = 1


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>LHS =\(\frac{1}{cos^2A}\) - \(\frac{1}{cot^2A}\) = sec<sup>2</sup>A -- tan<sup>2</sup>A = 1 + tan<sup>2</sup>A - tan<sup>2</sup>A</p> <p>=1 = RHS proved</p>

Q7:

Prove the following:

\(\frac{secA}{cosA}\) - \(\frac{tanA}{cotA}\) = 1


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>L.H.S. =\(\frac{secA}{cosA}\) - \(\frac{tanA}{cotA}\) = secA. \(\frac{1}{cosA}\) - tanA. \(\frac{1}{cotA}\)</p> <p>= secA. secA - tanA. tanA</p> <p>= sec<sup>2</sup>A - tan<sup>2</sup>A = 1 = R. H. S. Proved</p>

Q8:

Prove that:

\(\frac{1-tanA}{1+tanA}\) =\(\frac{cotA - 1}{cotA + 1}\)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>Here,\(\frac{1-tanA}{1+tanA}\) =\(\frac{\frac{1}{tanA}-\frac{tanA}{taanA}}{\frac{1}{tanA}+\frac{tanA}{tanA}}\)</p> <p> = \(\frac{cotA - 1}{cotA + 1}\) RHS proved</p>

Q9:

Prove that:

(tanθ + secθ)2 = \(\frac{1 + sinθ}{1 - sinθ}\)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p>
<p>L.H.S = ( tan&theta; + sec&theta;)<sup>2</sup></p>
<p>= (\(\frac{sin&theta;}{cos&theta;}\) + \(\frac{1}{cos&theta;}\))<sup>2</sup></p>
<p>= (\(\frac{1 + sin&theta;}{cos&theta;}\))<sup>2</sup></p>
<p>= \(\frac{(1 + sin&theta;)( 1 + sin&theta;)}{cos^2&theta;}\)</p>
<p>=\(\frac{(1 + sin&theta;)(1 + sin&theta;)}{1 - sin^2&theta;}\)</p>
<p>= \(\frac{1 + sin&theta;}{1 - sin&theta;}\) = RHS Proved.</p>

Q10:

Prove that:

\(\frac{1}{1 - cosA}\) - \(\frac{1}{1 + cosA}\) = 2cotA cosecA


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p>
<p>LHS. =\(\frac{1}{1 - cosA}\) - \(\frac{1}{1 + cosA}\) = \(\frac{1 + cosA -(1 - cosA)}{(1 - cosA) ( 1 + cosA)}\)</p>
<p>= \(\frac{1 + cosa - 1 + cosA}{1 - cos^2 A}\)</p>
<p>= \(\frac{2cosA}{sin^2 A}\) =\(\frac{2cosA}{sinA}\) .\(\frac{1}{sinA}\)</p>
<p>= 2cotA cosecA = RHS Proved.</p>

Q11:

Prove that:

\(\frac{cosx}{1 - sinx}\) + \(\frac{cosx}{1 + sinx)}\) = 2secx


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>Taking LHS,<br>=\(\frac{cosx}{1 - sinx}\) + \(\frac{cosx}{1 + sinx)}\)</p> <p>= \(\frac{cosx(1 + sinx) + cosx(1 - sinx)}{(1 - sinx) (1 + sinx)}\)</p> <p>= \(\frac{cosx + cosx.sinx + cosx - cosx.sinx}{1 - sin^2 x}\)</p> <p>= \(\frac{2cosx}{cos^2x}\) =\(\frac{2}{cosx}\) = 2secx = RHS proved. Ans</p>

Q12:

Prove that:

sin4θ + cos4θ = 1 - 2sin2θ cos2θ


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln: <br>Taking LHS,<br>=sin<sup>4</sup>&theta; + cos<sup>4</sup>&theta; = (sin<sup>2</sup>&theta;)<sup>2</sup> + (cos<sup>2</sup>&theta;)<sup>2</sup></p> <p>=(sin<sup>2</sup>&theta; + cos<sup>2</sup>&theta;)<sup>2</sup> - 2sin<sup>2</sup>&theta; cos<sup>2</sup>&theta;</p> <p>= 1 - 2sin<sup>2</sup>&theta; cos<sup>2</sup>&theta; = RHS Proved</p>

Q13:

Prove that:

(1 + sinA + cosA)2 = 2(1 + sinA) (1 + cosA)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>LHS = ( 1 + sinA + cosA)<sup>2</sup></p> <p> ={(1 + sinA) +cosA}<sup>2</sup></p> <p> = (1 + sinA)<sup>2</sup> + 2(1 + sinA) cosA + cos<sup>2</sup>A</p> <p> =(1 + sinA)<sup>2</sup> + 2( 1 + sinA) cosA + cos<sup>2</sup>A</p> <p> = ( 1+ sinA)<sup>2<em></em></sup>+ 2(1 + sinA) cosA + 1 - sin<sup>2</sup>A</p> <p> = ( 1+ sinA)<sup>2</sup> + 2(1 + sinA) cosA + (1 + sinA) (1 - sinA)</p> <p> =(1 + sinA) ( 1 + sinA + 2cosA + 1 -sinA)</p> <p> = ( 1 + sinA) ( 2 + 2 cosA)</p> <p> = 2(1 + sinA) (1 + cosA) = RHS Proved</p>

Q14:

Prove that:

\(\frac{cosA - sinA + 1}{cosA + sinA - 1}\) =\(\frac{1 + cosA}{sinA}\)


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>LHS =\(\frac{cosA - sinA + 1}{cosA + sinA - 1}\)</p> <p>=\(\frac{\frac{cosA - sinA + 1}{sinA}}{\frac{cosA + sinA - 1}{sinA}}\)</p> <p>=\(\frac{cotA - 1 + cosecA}{cotA + 1 - cosecA}\)</p> <p>= \(\frac{(cotA + cosecA) - (cosec^2 A - cot^2 A)}{cotA - cosecA + 1}\)</p> <p>=\(\frac{1 (cotA + cosecA) - (cosecA - cotA) (cosecA + cotA) }{cotA - cosecA + 1}\)</p> <p>= \(\frac{(cosecA + cotA) (1 - cosecA + cotA)}{cotA - cosecA + 1}\)</p> <p>= cosecA + cotA =\(\frac{1}{sinA}\) + \(\frac{cosA}{sinA}\)</p> <p>= \(\frac{1 + cosA}{sinA }\) =RHS proved.</p> <p></p>

Q15:

Prove that:

\(\frac{secA - tanA}{secA + tanA}\) = 1 - 2 secA tanA + 2 tan2A.

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>LHS =\(\frac{secA - tanA}{secA + tanA}\) =\(\frac{(secA - tanA)}{secA + tanA}\)&times; \(\frac{secA - tanA}{secA - tanA}\)</p> <p> = \(\frac{(secA - tanA)^2}{sec^2 A -tan^2A}\)</p> <p> =\(\frac{sec^2 A - 2secA tanA + tan^2A}{1}\)</p> <p> = 1 + tan<sup>2</sup>A - 2secA tanA + tan<sup>2</sup>A</p> <p> = 1 - 2secA tanA + 2 tan<sup>2</sup>A = RHS proved.</p>

Q16:

Prove that:

(secA + cosecA)2 = ( 1+ cotA)2 + (1 + tanA)2

                               


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Soln:</p> <p>RHS = 1 + 2cotA + cot<sup>2</sup>A + 1 + 2 tanA + tan<sup>2</sup>A</p> <p> = 1 + cot<sup>2</sup>A + 1 + tan<sup>2</sup>A + 2 (cotA + tanA)<sup>2</sup></p> <p><sup></sup>= 1 + cot<sup>2</sup>A + 1 + tan<sup>2</sup>A + 2 (cotA + tanA)</p> <p> = cosec<sup>2</sup>A + sec<sup>2</sup>A + 2(\(\frac{cosA}{sinA}\) + \(\frac{sinA}{cosA}\))</p> <p> =coscec<sup>2</sup>A +sec<sup>2</sup>A + 2(\(\frac{cos^2A + sin^2A}{sinA cosA}\))</p> <p> = cosec<sup>2</sup>A + sec<sup>2</sup>A + 2. \(\frac{1}{sinA cosA}\)</p> <p> = cosec<sup>2</sup>A + sec<sup>2</sup>A + 2coosecA secA</p> <p> = (secaA + cosecA)<sup>2</sup> =LHS proved</p> <p> </p>

Q17:

sec4θ - cosec4θ

 


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: Here, sec<sup>4</sup><em>&theta;</em> - cosec<sup>4</sup><em>&theta;</em></p> <p></p> <p> = (sec<sup>2</sup><em>&theta;</em>)<sup>2</sup> - (cosec<sup>2</sup><em>&theta;</em>)<sup>2</sup></p> <p> = ( sec<sup>2</sup><em>&theta; - cosec<sup>2</sup>&theta;) (sec<sup>2</sup>&theta; + cosec<sup>2</sup>&theta;)</em></p> <p><em> = ( sec&theta; - cosec&theta; ) (sec&theta; - cosec&theta; ) (sec<sup>2</sup>&theta; + cosec<sup>2</sup>&theta;)</em></p>

Q18:

sin2x + 3 sinx + 2


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: here, sin<sup>2</sup>x + 3sinx + 2</p> <p>= sin<sup>2</sup>x + 2sinx + sinx + 2</p> <p>= sinx (sinx + 2) +1 (sinx + 2)</p> <p>= (sinx + 2 ) (sinx + 1)</p>

Q19:

(1 - cos2  A) (1 + cot2 A)


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: L.H.S =(1 - cos<sup>2</sup> A ) (1 + cot<sup>2</sup> A ) = 1 &there4; 1 + cot<sup>2</sup>&theta; </p> <p> = sin<sup>2</sup> A . cosec<sup>2</sup> A = sin<sup>2</sup> A&times; \(\frac{1}{sin^2}A\) = cosec<sup>2 </sup>&theta; </p> <p> = 1 = R.H.S proved. and 1 - cos<sup>2</sup>&theta; = sin<sup>2</sup>&theta; </p> <p></p>

Q20:

tanθ . \(\sqrt{1}-{sin^2θ}\) = sinθ


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: L.H.S=tan&theta; . \(\sqrt{1} {sin^2&theta;}\) = tan&theta; \(\sqrt{cos^2&theta;}\)</p> <p> = tan&theta; . cos&theta; = \(\frac{sin&theta;}{cos&theta;}\).cos&theta;=sin&theta;= R.H.S. proved</p> <p></p>

Q21:

tan2 A - sin2 A = sin2 A . tan2 A


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: L.H.S = tan<sup>2</sup> A - sin<sup>2</sup> A = \(\frac{sin^2A}{cos^2A}\) - sin<sup>2</sup> A</p> <p> = \(\frac{sin^2 A- sin^2 A cos^2 A}{cos^2}\)=\(\frac{sin^2 (1 - cos^2 A)}{cos^2}\)</p> <p> = sin<sup>2</sup>A . \(\frac{sin^2A}{cos^2A}\)= sin<sup>2</sup> A. tan<sup>2</sup> A</p> <p> = R.H.S. proved.</p>

Q22:

\(\frac{1}{secA-tanA}\)-\(\frac{1}{cosA}\)= \(\frac{1}{cosA}\)- \(\frac{1}{secA+tanA}\)


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>soln; L.H.S=\(\frac{1}{secA-tanA}\)-\(\frac{1}{cosA}\)</p> <p> = \(\frac{sec^2A-tan^2A}{secA-tanA}\)-\(\frac{1}{cosA}\)</p> <p> = \(\frac{(secA+tanA)(secA-tanA}{(secA-tanA}\)-secA</p> <p> = secA - secA + tanA = tanA</p> <p> &there4; L.H.S.= R.H.S. proved.</p> <p>R.H.S = \(\frac{1}{cosA}\)-\(\frac{1}{secA + tanA}\)</p> <p> = secA-\(\frac{sec^2+tan^2}{secA + tanA}\)</p> <p> = sec A-\(\frac{(secA+tanA)(secA-tanA)}{(secA+tanA)}\)</p> <p> = secA - secA + tanA=tanA</p> <p>&there4; L.H.S=R.H.S. proved.</p>

Q23:

cosec4A (1-cos4A) = 1+2 cot2A


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>soln:<sup></sup>L.H.S. = cosec<sup>4</sup>A (1-cos<sup>4</sup>A)</p> <p> = cosec<sup>4</sup>A(1-cos<sup>2</sup> A) (1+cos<sup>2</sup> A)= cosec<sup>4</sup>&theta;[1-(cos<sup>2</sup>A)<sup>2</sup>]</p> <p> = cosec<sup>2</sup> A. cosec<sup>2</sup> A. sin<sup>2</sup> A (1+cos<sup>2</sup>A)</p> <p> = cosec<sup>2</sup>A.\(\frac{1}{sin^2A}\).sin<sup>2</sup>A(1+cos<sup>2</sup>A)</p> <p> = cosec<sup>2</sup>A (1+cos<sup>2</sup>A)</p> <p> = cosec<sup>2</sup>A+cosec<sup>2</sup>A cos<sup>2</sup>A= cosec<sup>2</sup>A + \(\frac{cos^2A}{sin^2A}\)</p> <p> = cosec<sup>2</sup>A + cot<sup>2</sup> A = 1 + cot<sup>2</sup>A+cot<sup>2</sup>A</p> <p> = 1+2 cot<sup>2</sup>A= R.H.S. proved.</p>

Q24:

sin3θ= 3 sinθ - 4 sin3θ


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: Given,&theta; = 30<sup>0</sup></p> <p> L.H.S. = sin3&theta; = sin 3&times; 30<sup>0</sup> = sin90<sup>0</sup> = 1</p> <p> R.H.S. = 3sin&theta; - 4sin<sup>3</sup>&theta; = 3sin30<sup>0</sup>- 4sin<sup>3</sup>30<sup>0</sup></p> <p> = 3&times;\(\frac{1}{2}\)-4 \(\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}\)<sup>3</sup> = \(\frac{3}{2}\)- \(\frac{4}{8}\)</p> <p> = \(\frac{3}{2}\)- \(\frac{1}{2}\)= \(\frac{3-1}{2}\)=\(\frac{2}{2}\)=1</p> <p> &there4; L.H.S.= R.H.S. proved.</p>

Q25:

sin (∝+ß) = sin∝cosß + cos ∝ sinß

 


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>soln: given,&prop;=0<sup>0</sup>,&beta;=30<sup>0</sup></p> <p>L.H.S.= sin(&prop; + &beta;) = sin (0<sup>0</sup> + 30<sup>0</sup>)= sin30<sup>0</sup>=\(\frac{1}{2}\)</p> <p>R.H.S = sin&prop; cos&beta; + cos&prop; sin&beta;</p> <p>= sin0<sup>0</sup> cos0<sup>0</sup> + cos0<sup>0</sup> sin0<sup>0</sup></p> <p>= 0&times;\(\frac{\sqrt{3}}{2}\)+ 1 \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) = 0 +\(\frac{1}{2}\)=\(\frac{1}{2}\)</p> <p>&there4; L.H.S = R.H.S. proved.</p>

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Gender Advocacy in Nepal

Gender Advocacy in Nepal

Gender advocacy refers to the discussion in various aspect based on gender in society. The discussion is mainly focused on the issues of male and female. It includes the role of male and female, social change, socialization, discrimination from gender perspective etc as the major aspects of the discussion. The kind of discussion and interaction is conducted with the aim of increasing awareness, empowering with rights and increasing the participation in various activities. These kinds of activities ultimately help and promote in the formation of gender related programmes, policies and laws and implement them effectively.Gender advocacy should be launched focusing on social inequality, events, social obstacles related to gender.

Identification of problem

First of all, gender issues for gender advocacy should be identified. It can be identified with the discussion of stakeholders. The depth of subject matter and impact to the society should be made clear. The issue for advocacy should be selected for the social welfare rather than personal welfare.

Creation of source

Various sources such as human, financial, physical facilities of advocacy what is its scope, how much resources are too launched and information related gender issues are to be disseminated in the magazine, poster, photograph, radio and television. The gender advocacy becomes highly successful if all resources are collected and mobilized on time.

Programme conduction

Program conduction is basic for advocacy program. Program should have definite purposeless and practical value. Various concerned persons and community should be involved in the programmes. Involvement cooperates to make the advocacy program successful.

Monitoring and evaluation

The most important thing is the impact of advocacy in the society. Monitoring and evaluation is necessary in order to identify the impact of the program. A survey can also be conducted. After monitoring, evaluation can be done to assess the impact of advocacy program on the target women. Next program can be made on the basic of the achievement made by the advocacy. Cooperation of all concerned can make gender advocacy highly successful.

Lesson

Causes and Effects of Population Change

Subject

Enviroment Population and Health

Grade

Grade 9

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